Quadratic equations

1. Factorising quadratic expressions

Factorising a trinomial

Factorise the following: 3x24x+1.

Answer: 3x24x+1=
polynomial
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Start by writing 3x24x+1=(x+)(x+). Choose any two factors of 3 that can be the coefficients of x and any two factors of 1 that can replace the 's such that the two expressions on either side of the '=' sign become equal.


STEP: Factorise 3x24x+1
[−2 points ⇒ 0 / 2 points left]
3x24x+1=(x1)(3x1)

The correct answer is: (x1)(3x1).


Submit your answer as:

Introduction to factorisation: fill in the blank

The equation here shows the factorisation of the expression x2+x. However, there is one value missing from the answer. What is the missing value?

x2+x=?_(x+1)
Answer:
expression
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]
You must find a number (or variable) which makes the equation true. Try putting a value into the blank and then distributing into the brackets to see if you get x2+x.
STEP: <no title>
[−1 point ⇒ 0 / 1 points left]

This question is about factorisation: we have the equation x2+x=?_(x+1) and we must fill in the missing value to make the equation true.

Factorisation is the inverse (the reverse) of the distributive law. When we factorise, we are "undoing" the distributive law. In other words, factorisation is when you "un-expand" or "undistribute" an expression. The picture below shows the relationship between distributing and factorising. When you multiply into the brackets you are distributing; when you put a number back outside of the brackets, you are factorising.

The equation given is: x2+x=?_(x+1). Imagine distributing the ? into the brackets: we need a number so that ?×x will be equal to x2. (It also must work out that ?×1 will be x.) This means that the missing value must be x.

Here are some important facts about factorisation:

  • the number of terms in the brackets is equal to the number of terms in the original expression
  • the value outside of the brackets is the HCF (Highest Common Factor) of the terms in x2+x
  • we can check the answer with the distributive law:
    x(x+1)bracketsExpand thex2+xoriginal expressionCool - this is the

The correct answer for the missing value is: x.


Submit your answer as:

Factorising: common factors

Factorise: 22yd42yb2+2yd6b3.

Answer:
polynomial
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

To factorise this expression we must take out the highest common factor (HCF) from all the terms.


STEP: Identify the highest common factor (HCF) of all the terms and factor it out
[−4 points ⇒ 0 / 4 points left]

To factorise this expression we must take out the highest common factor (HCF) from all the terms. In this case the HCF is 2y, so we get:

22yd42yb2+2yd6b3=2y(11d4b2+d6b3)

Submit your answer as:

Factorising ax2+bx+c if c<0

Factorise the following expression:

3x2x4
Answer: 3x2x4=
polynomial
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]
Since the constant term is negative, the factors will be something like (_+_)(__) or (__)(_+_).
STEP: Set up the binomials based on the 3x2 term and the signs
[−2 points ⇒ 2 / 4 points left]

The question asks you to factorise the expression 3x2x4. Factorising an expression usually requires patience, but in this case it is worse than usual because the quadratic coefficient is not 1: that adds to the complexity of the question.

Factorisation typically requires a trial and error approach, but you can usually use the quadratic term and the signs to start breaking down the factors. In this case:

  • the 3x2 term means that the factors must start with 3x and x
  • since the constant term is negative, the factors must have opposite signs (one positive and one negative)

That gives us two options at this point:

(3x+_)(x_)--OR--(3x_)(x+_)

Note: these are two possible answer for the question. In the first option you have 3x plus something while in the second you have 3x minus something; when you do FOIL with these two situations, you will not get the same answer. In other words, one of them will turn out to be right and the other will turn out to be wrong. Right now we do not know which will be correct.

(If you are not sure where these values came from, imagine starting FOIL with these two brackets and compare to the expression 3x2x4.)


STEP: Find the numbers which belong in the binomials
[−2 points ⇒ 0 / 4 points left]

Now we need to figure out which numbers belong in the empty spaces. Whatever numbers we try to put in the binomials, they should have a product of 4 so that we can get the last term in 3x2x4. (To be completely accurate, the numbers should have a product of 4, but we have already taken account of the negative in the signs in the binomials.)

For example, see what happens if you try the numbers 4 and 1. Remember that there are two options for where to put the signs in the binomials:

(3x+4)(x1)Try the values 4 and 1(3x4)(x+1)3x23x+4x43x2+3x4x43x2+x4Is either result correct?3x2x4

Nice! The answer on the right is the trinomial you needed. That means the binomials on the right side must be correct! It is quite nice that it worked out on the first try: it often takes more than one try.

The correct answer is: (3x4)(x+1).


Submit your answer as:

Prime expressions

Answer the two questions which follow about the expression 5x2+4

  1. What is the HCF (highest common factor) of the terms in the expression?

    Answer:
    expression
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]
    The highest common factor (HCF) of two things is the largest number (or biggest expression) that fits into both things. For example, the HCF of 8y2 and 12y is 4y because it fits into both of the expressions.
    STEP: <no title>
    [−1 point ⇒ 0 / 1 points left]

    We have the expression 5x2+4 and we need to find the HCF of the two terms in the expression. The HCF of the two terms is the largest expression (biggest number of factors) you can find which fits into both terms without a remainder.

    The terms 5x2 and 4 do not have any factors in common: they do not share factors of x, nor do they share any number factors. In a case like this, the largest common factor is 1.

    The HCF for the terms 5x2 and 4 is 1.


    Submit your answer as:
  2. Is the expression prime or not?

    Answer:
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]
    "Prime" means that something cannot be broken into smaller factors. For example, 7 is a prime number because it cannot be broken into smaller factors. Similarly, an expression is prime if it cannot be broken in to smaller factors.
    STEP: <no title>
    [−1 point ⇒ 0 / 1 points left]

    The question asks us to decide if the expression 5x2+4 is prime or not. "Prime" means that something cannot be broken down into smaller factors. For example, 10 is not prime because it can be broken down to 2×5; but 11 is prime because it cannot be broken into smaller factors: it can only be written as 1×11. The same idea also applies to expressions: an expression is prime if it cannot be broken into factors.

    How does the "prime" concept apply to the expression 5x2+4? From above we know that the HCF is 1: that means we cannot break up the expression more than this:

    5x2+41×(5x2+4)

    Therefore, the expression in this question is prime.

    The correct choice from the list is: It is prime.


    Submit your answer as:

Introduction to factorisation: multiple choice

For the expression 5x+5y, which choice below is the correctly factorised expression?

A 25(x+y)
B 5(x+5y)
C 5(x+y)
Answer:

The correctly factorised expression is: .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Use the distributive law to expand the expressions in the table. The correct choice must be equal to 5x+5y when it is expanded.


STEP: Check the answer choices by expanding them
[−1 point ⇒ 0 / 1 points left]

This question is about factorisation. We have the expression 5x+5y and we need to select the choice in the table which is the correctly factorised version of that expression.

Factorisation is the inverse (the reverse) of the distributive law. Factorising means "undoing" the distributive law. In other words, factorisation is when you "un-expand" or "undistribute" an expression. The picture below shows the relationship between distributing and factorising. When we multiply into the brackets we are distributing; when we put a number back outside of the brackets, we are factorising.

We can use the relationship above to find the correct answer to this question! Just start by expanding the first choice in the table with the distributive law to see if it is equal to 5x+5y:

Expand theexpression...25(x+y)...to see if youget 5x+5y25x+25y

Shucks: it did not work out correctly. We needed the expression 5x+5y but instead we got 25x+25y. Move on to checking the second choice in the table: expand it using the distributive law. Once you find the expression which is equal to 5x+5y when you expand it, you have the correct answer.

In this question the correct answer is 5(x+y) because when we distribute the 5 into the brackets we get 5x+5y.

The correct choice from the table is: C.


Submit your answer as:

Factorisation with trinomials

Factorise the expression given below.

INSTRUCTION: If it cannot be factorised, type "prime" in the answer box.
15x+12y+12z2
Answer:
polynomial
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]
You need to figure out the HCF of all three terms!
STEP: Determine the HCF (Highest Common Factor) of the terms in the expression
[−1 point ⇒ 2 / 3 points left]

You have the expression 15x+12y+12z2 and you must factorise it. Factorisation is the inverse of the distributive law, which means it is when you "un-expand" or "undistribute" an expression.

The first thing you need to do is find the HCF of the expression. However, the expression is a trinomial, so you must find the HCF of all three terms! The terms are 15x, 12y and 12z2. The HCF for these three terms is 3.


STEP: Pull out the HCF, and then use the distributive law to check your answer
[−2 points ⇒ 0 / 3 points left]

Once you know the HCF, you can factorise the expression. Start by putting the HCF in front of a pair of brackets, like this: 15x+12y+12z2=3×(more stuff). If you take 3 out of each term in the trinomial you will get 3(5x+4y+4z2).

Remember that you should always check your answer when you factorise. To check the answer, distribute the HCF back into the brackets to make sure that everything agrees with the original expression. Remember, you must distribute the 3 to all three terms in the brackets.

3(5x+4y+4z2)15x+12y+12z2original expressionAll right! this is the

The correct answer is: 3(5x+4y+4z2).


Submit your answer as:

Factorising x2+bx+c when c>0

Factorise the following expression into two binomials:

x2+10x+24

Your answer should look something like this: (x + 1)(x + 7).

Answer:
polynomial
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]
Start your answer by writing out: (x+_)(x+_). Then you must find two numbers which make the factorisation work out correctly.
STEP: Set up the binomials based on the x2 term
[−2 points ⇒ 2 / 4 points left]

We have the expression x2+10x+24 and we must factorise it.

To factorise an expression means to write the expression as a product. For example, if you factorise the number 10, you would write 25... 2 and 5 are factors of 10. In the case of a trinomial like x2+10x+24 factorising usually leads to two binomials: we want two binomials which have a product equal to x2+10x+24, just like 2 and 5 have a product of 10.

To work out the binomial factors requires patience, because there is no single calculation which will give us the answer. Rather we must take the time to work out the answer by trial and error (guess & check).

The first step is to set up the answer based on the x2 term in the trinomial. It tells us that the binomials must be like this:

(x+_)(x+_)

Imagine starting FOIL with these two brackets: you will get x2 from the two x's, which is the first part of x2+10x+24.


STEP: Find the numbers which belong in the binomials
[−2 points ⇒ 0 / 4 points left]

So far we have this: x2+10x+24=(x+_)(x+_). Now we need to figure out which numbers belong in the empty spaces. Whatever numbers we try to put in the binomials, we must check the binomials with FOIL to make sure that everything works out correctly.

We should try numbers which have a product of 24. This is because of the "Lasts" term in FOIL. For example, let's check what happens if we try the numbers 2 and 12:

(x+2)(x+12)expand the productUse FOIL tox2+12x+2x+24x2+14x+24trinomial we want?Is this the

Shucks - we did not get the trinomial we needed because the x term is not correct. That means that the binomials (x+2) and (x+12) cannot be the correct factors.

This is where patience is required: we need to try two other numbers which have a product of 24 and see if those numbers work out to make the trinomial we need. We must continue this process until we find two binomials which agree with the trinomial. The correct pair of binomials is (x+4)(x+6) because (x+4)(x+6)=x2+10x+24.

The correct answer is: (x+4)(x+6).


Submit your answer as:

Factorising a trinomial

Factorise the following:

q2+14q+40
Answer: q2+14q+40=
polynomial
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Start by writing q2+14q+40=(q+)(q+). Choose any two factors of +40 that can replace the 's such that the two expressions on either side of the '=' sign become equal.


STEP: Factorise q2+14q+40
[−2 points ⇒ 0 / 2 points left]

Factorisation usually requires patience: it is not something you can do by following a formula. So give yourself time to find the answer if it does not come quickly.

The most useful thing to look at to start factorisation is the constant term; in this case it is equal to +40. That means that when you factorise you must find two numbers which have a product of ('product' means multiply) +40. The bad news is that those two numbers must also agree with the other terms in the question, which often makes it a bit complex.

In this case it works out like this:

q2+14q+40=(q+4)(q+10)
NOTE: It is very important when you factorise to check your answer; you can do that by expanding the binomials (using FOIL). If you check and your answer does not work, make some changes and try again!

Submit your answer as:

Factorising quadratic trinomials

Factorise the following expression if possible. If not possible, type "prime" in the answer box.

9z235z+24
Answer:
polynomial
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]
Here are two hints:
  1. use the signs in the trinomial to figure out what the signs in the binomials should be
  2. be careful to conclude that the trinomial is prime if you try a few combinations and they all fail: you need to try all possible combinations of numbers to get the quadratic and the constant term. If none of those combinations work correctly, then the trinomial is prime

STEP: <no title>
[−4 points ⇒ 0 / 4 points left]

This question is tough: both the quadratic coefficient and the constant term are composite numbers (not prime), which means there are a lot of combinations of numbers that you may need to try to get the answer. Ugh.

Use the signs in the trinomial as a guide for the signs in the binomials: since the constant term is positive but the linear term is negative, the signs in the binomials must both be negatives. Once you know what to expect with the signs, start working with the numbers to try to find the factors.

Based on the trinomial, here is one possible starting point:

(9z_)(z_)

If this option does not lead to the correct factors, you may have to come back to this starting point and make a change. It is possible that the trinomial is prime, but remember that you must try all possible combinations of numbers to get the quadratic and the constant term from the trinomial; if none of those combinations work correctly, then the trinomial is prime.

The correct factors are (9z8)(z3), because when you expand the brackets it works out:

(9z8)(z3)=9z227z8z+24=9z235z+24

The correct answer is: (9z8)(z3).


Submit your answer as:

Factorising quadratic expressions

Answer the following two questions about the expression f210f+25.

  1. Factorise the expression. If the expression cannot be factorised, type "prime" in the answer box.

    Answer:
    polynomial
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    STEP: Check to see if the expression is a special pattern
    [−1 point ⇒ 2 / 3 points left]

    The question gives the expression f210f+25 and you must factorise it.

    The first thing you should do is check to see if the expression is either a difference of two squares or the square of a binomial. A difference of two squares must have two terms made of squared values, with a subtraction between them. For a square of a binomial, the coefficient of the middle term must be twice as much as the square root of the constant term.

    The expression cannot be a difference of two squares: it has too many terms (a difference of two squares is always a binomial). However, the expression does match the pattern for the square of a binomial: the constant term is the square of 5 while the coefficient of the middle term is 2(5).

    This question gives an expression with the square of a binomial pattern.


    STEP: Find the binomials using the pattern
    [−2 points ⇒ 0 / 3 points left]

    The square of a binomial pattern grows from the square root of the constant term (which is also half of the coefficient of the linear term). In this case, that number is 5:

    f210f+25=(f5)(f5)

    Remember that it is always a good idea to check your binomial factors with FOIL in order to make sure that everything works out: it is critical that you get it exactly right!

    The correct binomial factors are: (f5)(f5). You can also write the answer as (f5)2.


    Submit your answer as:
  2. Which of the following choices completes the statement correctly about the expression?

    Answer: The signs in the binomials must be .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]
    The correct answer is related to the rules for multiplication of positives and negatives.
    STEP: <no title>
    [−1 point ⇒ 0 / 1 points left]

    This question is about the relationship between the signs in the binomials and the signs in the expression f210f+25. These signs are related to each other by the rules for multiplying positives and negatives, and the signs "meet" (multiply each other) according to the connections in FOIL.

    In this question, the constant term is 25. Since this number comes from a multiplication during the FOIL calculations, the rules for positives and negatives with multiplication are at work. Therefore, the numbers that create the 25 must have the same type of signs (either positive-positive or negative-negative) because when you multiply numbers with the same signs, you get a positive answer, like 25 in the expression f210f+25.

    The correct ending to the sentence is that the binomials must have: the same signs because the constant term is positive.


    Submit your answer as:

Factorising x2+bx+c when c<0

Factorise the following expression into two binomials:

x26x7

Your answer should look something like this: (x + 1)(x - 5).

Answer:
polynomial
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]
Start your answer by writing out: (x+_)(x_). The signs in the binomials must be opposites - think about why! Then you must find two numbers which make the factorisation work out correctly.
STEP: Set up the binomials based on the x2 term
[−2 points ⇒ 2 / 4 points left]

You have the expression x26x7 and you must factorise it. To work out the binomial factors often requires patience: there is no single calculation which will give you the answer. Rather you must take the time to work out the answer by trial and error (guess & check).

For the expression x26x7, the first and last terms both give big hints about the binomials that you need. The first term suggests that the factors must both have x in them: (x)(x).

The last term in x26x7 tells you about the signs in the binomials. Remember that the number 7 usually comes from multiplying the numbers in the binomials (when you do FOIL). Therefore, the binomials must contain one negative sign and one positive sign. Therefore the binomials must be:

(x+_)(x_)

So at this point we have x26x7=(x+_)(x_).


STEP: Find the numbers which belong in the binomials
[−2 points ⇒ 0 / 4 points left]

Now you need to figure out which numbers belong in the empty spaces. Whatever numbers you try to put in the binomials, you must check the binomials with FOIL to make sure that everything works out correctly.

You should try numbers which have a product of 7. For example, let's check what happens with the numbers 1 and 7:

(x+1)(x7)expand the productUse FOIL tox27x+x7x26x7trinomial we want?Is this the

Nice! This is the trinomial you needed, so the binomials must be correct!

The correct answer is: (x+1)(x7). You can also write the answer as (x7)(x+1).


Submit your answer as:

Factorisation: the distributive law backwards

Factorise the following expression by taking the HCF (Highest Common Factor) of the terms outside of brackets:

2x2+x

Your answer should look something like this: 3(x + 7).

Answer:
polynomial
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]
You need to figure out the HCF of the two terms first; then use the HCF to get the answer.
STEP: Determine the HCF (Highest Common Factor) of the terms in the expression
[−1 point ⇒ 2 / 3 points left]

We have the expression 2x2+x and we must factorise it. Factorisation means that we need to "undo" the distributive law - factorisation is the inverse of distributing. In other words, factorisation is when you "un-expand" or "undistribute" an expression.

The first thing we need to do is find the HCF of the two terms in the expression. (Remember that the HCF is the largest number or value which fits into both of the terms.) For the expression 2x2+x the HCF of the terms is x.


STEP: Pull out the HCF, and then use the distributive law to check your answer
[−2 points ⇒ 0 / 3 points left]

Once you know the HCF, you can factorise the expression. As mentioned above, factorisation is when we "undo" the distributive law. The picture below shows the relationship between distributing and factorising. When you multiply into the brackets you are distributing; when you put a number back outside of the brackets, you are factorising.

In this question we are factorising, so start by putting the HCF in front of a pair of brackets, like this: 2x2+x=x(more stuff). Then you must fill in the brackets so that the expressions are equal.

In this case, if you take x out of the terms in the expression you will get x(2x+1). Can you see why factorisation means "undistributing"?

Checking the answer: When we factorise, we should always take the time necessary to check the answer. To check the answer we need to distribute the HCF, x, back into the brackets to make sure that everything agrees with the original expression.

x(2x+1)back into the bracketsMultiply the HCF2x2+xoriginal expressionCool - this is the

The correct answer is: x(2x+1).


Submit your answer as:

More complex HCFs

Factorise the following expression.

INSTRUCTION: If it cannot be factorised, type "prime" in the answer box.
6a4+5
Answer:
string
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]
You need to figure out the HCF of the two terms first; then use the HCF to get the answer.
STEP: Determine the HCF (Highest Common Factor) of the terms in the expression
[−1 point ⇒ 2 / 3 points left]

You have the expression 6a4+5 and you must factorise it. Factorisation is the inverse of the distributive law, which means it is when you "un-expand" or "undistribute" an expression.

The first thing you need to do is find the HCF of the two terms in the expression. The two terms are 6a4 and 5. These two terms have nothing in common at all, so the HCF is 1.


STEP: Pull out the HCF, and then use the distributive law to check your answer
[−2 points ⇒ 0 / 3 points left]

Once you know the HCF, you can factorise the expression. However, this is a special situation: the HCF is 1. That means that there are no common factors in the terms of the expression. Therefore, the expression must be prime!

A prime quantity has only two factors: 1 and itself. This expression has no other factors, only the HCF of 1 and the expression itself.

The correct answer is: prime.


Submit your answer as:

Factorising a difference of two squares

Factorise the following expression into two binomials.

4x225
Answer: 4x225=
polynomial
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]
The expression 4x225 has a special pattern: it is a difference of two squares. You can use this pattern to get to the answer quickly.
STEP: <no title>
[−2 points ⇒ 0 / 2 points left]

You need to factorise the expression 4x225. To work out the binomial factors often requires patience: however, there is a pattern in this expression which lets you get straight to the answer!

The expression 4x225 is a difference of two squares. (Both terms are squared values, and the subtraction is a difference.) The difference of two squares pattern is linked directly to binomials which are identical except for the signs in the binomials:

4x225=(2x+5)(2x5)
NOTE: You can check your answer by multiplying out:
(2x+5)(2x5)=x210x+10x25=4x225

The correct answer is: (2x+5)(2x5).


Submit your answer as:

Factorisation

Factorise the expression given below if possible. Type "prime" in the answer box if it cannot be factorised.

4x3y+20x2y220x2y
Answer:
polynomial
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]
You need to figure out the HCF of all three terms in the expression!
STEP: <no title>
[−1 point ⇒ 3 / 4 points left]

You must factorise the expression 4x3y+20x2y220x2y. Start by figuring out the HCF of the expression. When dealing with a trinomial, the HCF must come from all three of the terms in the expression.

The HCF for these terms is 4x2y.


STEP: <no title>
[−3 points ⇒ 0 / 4 points left]

Now factorise the HCF out of the expression. If you take 4x2y out of each term in the expression you will get 4x2y(x+5y5).

Finally, check your answer. Do this by distributing the HCF back into the brackets to make sure you get the original expression. Remember, you must distribute the 4x2y to each of the terms in the brackets.

4x2y(x+5y5) 4x3y+20x2y220x2y

The correct answer is: 4x2y(x+5y5).


Submit your answer as:

Factorisation: common factors

Factorise: 4d(s+8)+9(s+8).

Answer:
polynomial
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Take out the common factor which is the expression in brackets.


STEP: Identify the common factor and factorise
[−2 points ⇒ 0 / 2 points left]

We take out the common bracket from the two terms to get:

4d(s+8)+9(s+8)=(s+8)(4d+9)

The order of the brackets does not matter, since a×b is the same as b×a. Therefore the following is also correct:

4d(s+8)+9(s+8)=(4d+9)(s+8)

Submit your answer as:

Factorising quadratic trinomials

Factorise the following expression if possible. If the expression cannot be factorised, write "prime" in the answer box.

2x2+8x+9
Answer: 2x2+8x+9=
string
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]
Here are two things to keep in mind:
  1. All of the terms are positive, which means that the binomials must have positive signs in them: (_+_)(_+_)
  2. It is possible that the trinomial is prime; you need to try all possible combinations of numbers to get the quadratic and the constant term; if none of those combinations work correctly, then the trinomial is prime

STEP: Set up the binomials based on the 2x2 term and the signs
[−1 point ⇒ 2 / 3 points left]

You have the expression 2x2+8x+9 and you must factorise it. This question also has the challenging possibility that the expression is prime. To determine if the expression is prime, you must try all possible combinations of numbers to get the quadratic and the constant term from the trinomial; if none of those combinations work correctly, then the trinomial is prime.

For the factors of 2x2+8x+9, the following things must be true:

  • the 2x2 term means that the factors must start with 2x and x
  • since all of the terms in the trinomial are positive, the factors must both have plus signs

From that information, here is an intelligent starting point:

(2x+_)(x+_)


STEP: Find the numbers which belong in the binomials
[−2 points ⇒ 0 / 3 points left]

At this point you must try some numbers in the binomials; the numbers should have a product of 9. For example, see what happens if you try the numbers 3 and 3.

(2x+3)(x+3)values 3 and 3Try out the2x2+6x+3x+92x2+9x+9we want or not?Is this the trinomial

This is not the trinomial you needed because the x-term is not correct. That means that the binomials (2x+3) and (x+3) cannot be the correct factors.

Now you must try another combination of numbers in the binomials. For example, you can try a different pair of numbers which have a product of 9. For example, (2x+1)(x+9).. In the end, though, there are no combinations of numbers which agree with everything in the trinomial. This expressoin is prime.

The correct answer is: prime.


Submit your answer as:

Factorisation: common factors

Factorise the following:

6qf+66fd
Answer:6qf+66fd=
polynomial
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Start by identifying the highest common factor (HCF) of both terms.


STEP: Identify the highest common factor (HCF) of both terms and factorise
[−2 points ⇒ 0 / 2 points left]

We take out the highest common factor (HCF) from both terms. In this case, the HCF is 6f.

6qf+66fd=6f(q+11d)

When you have factorised the HCF out of the expression, there cannot be any more common factors remaining in the brackets!


Submit your answer as:

Factorising complex differences of two squares

Factorise the following expression into two binomials, if possible.

INSTRUCTION: If the expression cannot be factorised, type "prime" in the answer box.
36g249
Answer:
polynomial
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]
What is the pattern in the binomials for a difference of two squares?
STEP: <no title>
[−2 points ⇒ 0 / 2 points left]

You need to factorise the expression 36g249. Notice that all of the values in the expression are squared values (either square numbers or the square of a variable). Furthermore, the expression shows a difference (subtraction). All those facts together mean that the expression is a difference of two squares.

For the difference of two squares pattern, the binomials are always identical except for the signs in the binomials:

36g249=(6g+7)(6g7)

As always, you should check the binomials to make sure you get the exact expression you need.

(6g+7)(6g7)expand the productUse FOIL to36g242g+42g4936g249terms cancel each otherThe two middle

The correct answer is: (6g+7)(6g7). The order of the binomials is not important, so you can also write the answer as (6g7)(6g+7).


Submit your answer as:

Factorising x2+bx+c: true or false

The equation below shows a trinomial factorised into two binomial factors. Are the binomial factors correct or not? (Is the equation true or false?)

x2+4x12=(x18)(x+6)
Answer:
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]
Expand the binomials (with FOIL) to see if the you get the trinomial or not. Be careful to pay attention to the signs in the binomial.
STEP: <no title>
[−1 point ⇒ 0 / 1 points left]

Factorisation is the inverse of expanding something. When we factorise, we must break something apart into a product. For example, if you factorise the number 10, you write 25. For a trinomial, it is common that the factors are binomials, like what is shown in this question.

To see if the binomials are the correct factors, we must expand the product to find out if they are equal to the trinomial x2+4x12 or not. That means using FOIL:

(x18)(x+6)expand the productUse FOIL tox2+6x18x108x212x108trinomial we want?Is this the

We were supposed to get the trinomial x2+4x12! However, the constant term and the x-term are both wrong. Therefore x2+4x12 is not equal to (x+6)(x18)

Notice that we checked the answer by expanding the binomials with FOIL and comparing the result to the trinomial in the question. Checking the answer with FOIL is a normal part of factorisation: you will do it over and over again when you are solving factorisation problems.

The correct choice is: False.


Submit your answer as:

Factorising ax2+bx+c

Factorise the following expression into two binomials:

3x2+8x+4
Answer: 3x2+8x+4=
polynomial
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]
Since all of the terms in the trinomial are positive, the factors will be something like (_+_)(_+_).
STEP: Set up the binomials based on the 3x2 term and the signs
[−2 points ⇒ 2 / 4 points left]

You have the expression 3x2+8x+4 and you must factorise it. Factorising an expression usually requires patience, but in this case it is worse than usual because the quadratic coefficient is not 1: that adds to the complexity of the question.

Factorisation usually requires some trial and error, but you can usually use the quadratic term and the signs to start breaking down the factors. In this case:

  • the 3x2 term means that the factors must start with 3x and x
  • since all of the terms in the trinomial are positive, the factors must both have plus signs

That gives us:

(3x+_)(x+_)

(If you are not sure where these values came from, imagine starting FOIL with these two brackets and compare to the expression 3x2+8x+4.)


STEP: Find the numbers which belong in the binomials
[−2 points ⇒ 0 / 4 points left]

Now we need to figure out which numbers belong in the empty spaces. Whatever numbers we try to put in the binomials, they should have a product of 4 so that we can get the last term in 3x2+8x+4.

For example, let's check what happens if we try the numbers 2 and 2:

(3x+2)(x+2)values 2 and 2Try out the3x2+6x+2x+43x2+8x+4we want or not?Is this the trinomial

Nice! We got the trinomial we needed, so the binomials must be correct! It is quite nice that it worked out on the first try: it often takes more than one try.

The correct answer is: (3x+2)(x+2).


Submit your answer as:

Factorising quadratic trinomials: increasing complexity

Factorise the following expression. (It is possible to factorise the expression - it is not prime.)

14x2+x3
Answer:
polynomial
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Here are two things to keep in mind: Look at the constant term in the trinomial: it is negative, so the binomials must have opposite signs, like this: (_+_)(__)


STEP: Set up the binomials based on the 14x2 term and the signs
[−2 points ⇒ 2 / 4 points left]

For this question you must factorise the expression 14x2+x3. For the factors of 14x2+x3, the following things must be true:

  • the coefficient of the quadratic term is a composite number, so there is more than one possibility for how to break it down in the binomial factors; the correct values might be 14x and x, or they could be 7x and 2x
  • since the constant term is negative, the signs in the binomials must be opposites (one positive and one negative)

From that information, here is an intelligent starting point.

(14x+_)(x_)

This is a good start, but keep in mind that it may prove to be wrong - it is not the only possibility; for example, the signs can be swapped, which would give a different answer.


STEP: Try some numbers in the binomials
[−2 points ⇒ 0 / 4 points left]

Now try some numbers in the binomials to get the constant term of 3. For example, see what happens if you try the numbers 1 and 3.

(7x+1)(2x3)values 1 and 3Try out the14x221x+2x314x219x3we want or not?Is this the trinomial

Unfortunately, those are not the correct factors. If there are any other possible combinations of numbers to put into the binomials, you must try them to see which combination works out. For example, you can try swapping the signs in the binomials: (7x1)(2x+3).

The correct answer is: (7x3)(2x+1).


Submit your answer as:

Factorising x2+bx+c: filling in the blank

The equation below shows an incomplete factorisation. What is the missing number?

x22x15=(x5)(x+?_)
Answer:
numeric
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]
Use FOIL on the binomials to see if they are equal. Be careful to pay attention to the signs in the binomial.
STEP: <no title>
[−1 point ⇒ 0 / 1 points left]

We have the equation x22x15=(x5)(x+?_), which shows an unfinished factorisation question. We must complete the factorisation by filling in the missing number.

To factorise an expression means to write the expression as a product. Factorising the trinomial x22x15 leads to two binomials: (x5)(x+?_). We need to put a number into the blank space so that the binomials have a product equal to x22x15.

We must think about FOIL when we factorise into binomials: if we expand the binomials using FOIL we must get the trinomial x22x15 (because the product must be equal to that trinomial).

(x5)(x+?_)x22x15

Imagine doing FOIL with these two brackets: the "Lasts" term must be equal to 15! What number should go into the blank space for that to work out?

Notice that the constant term in the trinomial is negative. That is why the signs in the binomials are opposites: negative multiplied by positive is negative! 5×3=15 .

The complete factorised expression is (x5)(x+3), and the answer is: 3.


Submit your answer as:

Exercises

Factorising a trinomial

Factorise the following: 3t210t+8.

Answer: 3t210t+8=
polynomial
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Start by writing 3t210t+8=(t+)(t+). Choose any two factors of 3 that can be the coefficients of t and any two factors of 8 that can replace the 's such that the two expressions on either side of the '=' sign become equal.


STEP: Factorise 3t210t+8
[−2 points ⇒ 0 / 2 points left]
3t210t+8=(t2)(3t4)

The correct answer is: (t2)(3t4).


Submit your answer as:

Factorising a trinomial

Factorise: 9p29p+2.

Answer: 9p29p+2=
polynomial
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Start by writing 9p29p+2=(p+)(p+). Choose any two factors of 9 that can be the coefficients of p and any two factors of 2 that can replace the 's such that the two expressions on either side of the '=' sign become equal.


STEP: Factorise 9p29p+2
[−2 points ⇒ 0 / 2 points left]
9p29p+2=(3p1)(3p2)

The correct answer is: (3p1)(3p2).


Submit your answer as:

Factorising a trinomial

Factorise: 2y2+y3.

Answer: 2y2+y3=
polynomial
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Start by writing 2y2+y3=(y+)(y+). Choose any two factors of 2 that can be the coefficients of y and any two factors of 3 that can replace the 's such that the two expressions on either side of the '=' sign become equal.


STEP: Factorise 2y2+y3
[−2 points ⇒ 0 / 2 points left]
2y2+y3=(y1)(2y+3)

The correct answer is: (y1)(2y+3).


Submit your answer as:

Introduction to factorisation: fill in the blank

The equation here shows the factorisation of the expression x24x. However, there is one value missing from the answer. What is the missing value?

x24x=?_(x4)
Answer:
expression
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]
You must find a number (or variable) which makes the equation true. Try putting a value into the blank and then distributing into the brackets to see if you get x24x.
STEP: <no title>
[−1 point ⇒ 0 / 1 points left]

This question is about factorisation: we have the equation x24x=?_(x4) and we must fill in the missing value to make the equation true.

Factorisation is the inverse (the reverse) of the distributive law. When we factorise, we are "undoing" the distributive law. In other words, factorisation is when you "un-expand" or "undistribute" an expression. The picture below shows the relationship between distributing and factorising. When you multiply into the brackets you are distributing; when you put a number back outside of the brackets, you are factorising.

The equation given is: x24x=?_(x4). Imagine distributing the ? into the brackets: we need a number so that ?×x will be equal to x2. (It also must work out that ?×4 will be 4x.) This means that the missing value must be x.

Here are some important facts about factorisation:

  • the number of terms in the brackets is equal to the number of terms in the original expression
  • the value outside of the brackets is the HCF (Highest Common Factor) of the terms in x24x
  • we can check the answer with the distributive law:
    x(x4)bracketsExpand thex24xoriginal expressionCool - this is the

The correct answer for the missing value is: x.


Submit your answer as:

Introduction to factorisation: fill in the blank

The equation here shows the factorisation of the expression 6x18. However, there is one value missing from the answer. What is the missing value?

6x18=?_(x3)
Answer:
expression
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]
You must find a number (or variable) which makes the equation true. Try putting a value into the blank and then distributing into the brackets to see if you get 6x18.
STEP: <no title>
[−1 point ⇒ 0 / 1 points left]

This question is about factorisation: we have the equation 6x18=?_(x3) and we must fill in the missing value to make the equation true.

Factorisation is the inverse (the reverse) of the distributive law. When we factorise, we are "undoing" the distributive law. In other words, factorisation is when you "un-expand" or "undistribute" an expression. The picture below shows the relationship between distributing and factorising. When you multiply into the brackets you are distributing; when you put a number back outside of the brackets, you are factorising.

The equation given is: 6x18=?_(x3). Imagine distributing the ? into the brackets: we need a number so that ?×x will be equal to 6x. (It also must work out that ?×3 will be 18.) This means that the missing value must be 6.

Here are some important facts about factorisation:

  • the number of terms in the brackets is equal to the number of terms in the original expression
  • the value outside of the brackets is the HCF (Highest Common Factor) of the terms in 6x18
  • we can check the answer with the distributive law:
    6(x3)bracketsExpand the6x18original expressionCool - this is the

The correct answer for the missing value is: 6.


Submit your answer as:

Introduction to factorisation: fill in the blank

The equation here shows the factorisation of the expression x2+4x. However, there is one value missing from the answer. What is the missing value?

x2+4x=?_(x+4)
Answer:
expression
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]
You must find a number (or variable) which makes the equation true. Try putting a value into the blank and then distributing into the brackets to see if you get x2+4x.
STEP: <no title>
[−1 point ⇒ 0 / 1 points left]

This question is about factorisation: we have the equation x2+4x=?_(x+4) and we must fill in the missing value to make the equation true.

Factorisation is the inverse (the reverse) of the distributive law. When we factorise, we are "undoing" the distributive law. In other words, factorisation is when you "un-expand" or "undistribute" an expression. The picture below shows the relationship between distributing and factorising. When you multiply into the brackets you are distributing; when you put a number back outside of the brackets, you are factorising.

The equation given is: x2+4x=?_(x+4). Imagine distributing the ? into the brackets: we need a number so that ?×x will be equal to x2. (It also must work out that ?×4 will be 4x.) This means that the missing value must be x.

Here are some important facts about factorisation:

  • the number of terms in the brackets is equal to the number of terms in the original expression
  • the value outside of the brackets is the HCF (Highest Common Factor) of the terms in x2+4x
  • we can check the answer with the distributive law:
    x(x+4)bracketsExpand thex2+4xoriginal expressionCool - this is the

The correct answer for the missing value is: x.


Submit your answer as:

Factorising: common factors

Factorise the following: 4ns4+10nt510ns4t2.

Answer:
polynomial
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

To factorise this expression we must take out the highest common factor (HCF) from all the terms.


STEP: Identify the highest common factor (HCF) of all the terms and factor it out
[−4 points ⇒ 0 / 4 points left]

To factorise this expression we must take out the highest common factor (HCF) from all the terms. In this case the HCF is 2n, so we get:

4ns4+10nt510ns4t2=2n(2s4+5t55s4t2)

Submit your answer as:

Factorising: common factors

Factorise the following: 40pa2+8pw4+16pa4w2.

Answer:
polynomial
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

To factorise this expression we must take out the highest common factor (HCF) from all the terms.


STEP: Identify the highest common factor (HCF) of all the terms and factor it out
[−4 points ⇒ 0 / 4 points left]

To factorise this expression we must take out the highest common factor (HCF) from all the terms. In this case the HCF is 8p, so we get:

40pa2+8pw4+16pa4w2=8p(5a2+w4+2a4w2)

Submit your answer as:

Factorising: common factors

Factorise: 12fz5+24fj2+12fz4j4.

Answer:
polynomial
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

To factorise this expression we must take out the highest common factor (HCF) from all the terms.


STEP: Identify the highest common factor (HCF) of all the terms and factor it out
[−4 points ⇒ 0 / 4 points left]

To factorise this expression we must take out the highest common factor (HCF) from all the terms. In this case the HCF is 12f, so we get:

12fz5+24fj2+12fz4j4=12f(z5+2j2+z4j4)

Submit your answer as:

Factorising ax2+bx+c if c<0

Factorise the following expression:

3x28x3
Answer: 3x28x3=
polynomial
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]
Since the constant term is negative, the factors will be something like (_+_)(__) or (__)(_+_).
STEP: Set up the binomials based on the 3x2 term and the signs
[−2 points ⇒ 2 / 4 points left]

The question asks you to factorise the expression 3x28x3. Factorising an expression usually requires patience, but in this case it is worse than usual because the quadratic coefficient is not 1: that adds to the complexity of the question.

Factorisation typically requires a trial and error approach, but you can usually use the quadratic term and the signs to start breaking down the factors. In this case:

  • the 3x2 term means that the factors must start with 3x and x
  • since the constant term is negative, the factors must have opposite signs (one positive and one negative)

That gives us two options at this point:

(3x+_)(x_)--OR--(3x_)(x+_)

Note: these are two possible answer for the question. In the first option you have 3x plus something while in the second you have 3x minus something; when you do FOIL with these two situations, you will not get the same answer. In other words, one of them will turn out to be right and the other will turn out to be wrong. Right now we do not know which will be correct.

(If you are not sure where these values came from, imagine starting FOIL with these two brackets and compare to the expression 3x28x3.)


STEP: Find the numbers which belong in the binomials
[−2 points ⇒ 0 / 4 points left]

Now we need to figure out which numbers belong in the empty spaces. Whatever numbers we try to put in the binomials, they should have a product of 3 so that we can get the last term in 3x28x3. (To be completely accurate, the numbers should have a product of 3, but we have already taken account of the negative in the signs in the binomials.)

For example, see what happens if you try the numbers 3 and 1. Remember that there are two options for where to put the signs in the binomials:

(3x+3)(x1)Try the values 3 and 1(3x3)(x+1)3x23x+3x33x2+3x3x33x23Is either result correct?3x23

Shucks - neither result is equal to 3x28x3 (the x-terms are wrong in both options). That means that neither of the binomial options above is correct. Shame, you need to try another combination of numbers which have a product of 3; and remember that there are two ways the signs can sit in the binomials.

The correct answer is: (3x+1)(x3).


Submit your answer as:

Factorising ax2+bx+c if c<0

Factorise the following expression:

3x2+x2
Answer: 3x2+x2=
polynomial
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]
Since the constant term is negative, the factors will be something like (_+_)(__) or (__)(_+_).
STEP: Set up the binomials based on the 3x2 term and the signs
[−2 points ⇒ 2 / 4 points left]

The question asks you to factorise the expression 3x2+x2. Factorising an expression usually requires patience, but in this case it is worse than usual because the quadratic coefficient is not 1: that adds to the complexity of the question.

Factorisation typically requires a trial and error approach, but you can usually use the quadratic term and the signs to start breaking down the factors. In this case:

  • the 3x2 term means that the factors must start with 3x and x
  • since the constant term is negative, the factors must have opposite signs (one positive and one negative)

That gives us two options at this point:

(3x+_)(x_)--OR--(3x_)(x+_)

Note: these are two possible answer for the question. In the first option you have 3x plus something while in the second you have 3x minus something; when you do FOIL with these two situations, you will not get the same answer. In other words, one of them will turn out to be right and the other will turn out to be wrong. Right now we do not know which will be correct.

(If you are not sure where these values came from, imagine starting FOIL with these two brackets and compare to the expression 3x2+x2.)


STEP: Find the numbers which belong in the binomials
[−2 points ⇒ 0 / 4 points left]

Now we need to figure out which numbers belong in the empty spaces. Whatever numbers we try to put in the binomials, they should have a product of 2 so that we can get the last term in 3x2+x2. (To be completely accurate, the numbers should have a product of 2, but we have already taken account of the negative in the signs in the binomials.)

For example, see what happens if you try the numbers 2 and 1. Remember that there are two options for where to put the signs in the binomials:

(3x+2)(x1)Try the values 2 and 1(3x2)(x+1)3x23x+2x23x2+3x2x23x2x2Is either result correct?3x2+x2

Nice! The answer on the right is the trinomial you needed. That means the binomials on the right side must be correct! It is quite nice that it worked out on the first try: it often takes more than one try.

The correct answer is: (3x2)(x+1).


Submit your answer as:

Factorising ax2+bx+c if c<0

Factorise the following expression:

2x2x6
Answer: 2x2x6=
polynomial
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]
Since the constant term is negative, the factors will be something like (_+_)(__) or (__)(_+_).
STEP: Set up the binomials based on the 2x2 term and the signs
[−2 points ⇒ 2 / 4 points left]

The question asks you to factorise the expression 2x2x6. Factorising an expression usually requires patience, but in this case it is worse than usual because the quadratic coefficient is not 1: that adds to the complexity of the question.

Factorisation typically requires a trial and error approach, but you can usually use the quadratic term and the signs to start breaking down the factors. In this case:

  • the 2x2 term means that the factors must start with 2x and x
  • since the constant term is negative, the factors must have opposite signs (one positive and one negative)

That gives us two options at this point:

(2x+_)(x_)--OR--(2x_)(x+_)

Note: these are two possible answer for the question. In the first option you have 2x plus something while in the second you have 2x minus something; when you do FOIL with these two situations, you will not get the same answer. In other words, one of them will turn out to be right and the other will turn out to be wrong. Right now we do not know which will be correct.

(If you are not sure where these values came from, imagine starting FOIL with these two brackets and compare to the expression 2x2x6.)


STEP: Find the numbers which belong in the binomials
[−2 points ⇒ 0 / 4 points left]

Now we need to figure out which numbers belong in the empty spaces. Whatever numbers we try to put in the binomials, they should have a product of 6 so that we can get the last term in 2x2x6. (To be completely accurate, the numbers should have a product of 6, but we have already taken account of the negative in the signs in the binomials.)

For example, see what happens if you try the numbers 1 and 6. Remember that there are two options for where to put the signs in the binomials:

(2x+1)(x6)Try the values 1 and 6(2x1)(x+6)2x212x+x62x2+12xx62x211x6Is either result correct?2x2+11x6

Shucks - neither result is equal to 2x2x6 (the x-terms are wrong in both options). That means that neither of the binomial options above is correct. Shame, you need to try another combination of numbers which have a product of 6; and remember that there are two ways the signs can sit in the binomials.

The correct answer is: (2x+3)(x2).


Submit your answer as:

Prime expressions

Answer the two questions which follow about the expression x+5y

  1. What is the HCF (highest common factor) of the terms in the expression?

    Answer:
    expression
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]
    The highest common factor (HCF) of two things is the largest number (or biggest expression) that fits into both things. For example, the HCF of 8y2 and 12y is 4y because it fits into both of the expressions.
    STEP: <no title>
    [−1 point ⇒ 0 / 1 points left]

    We have the expression x+5y and we need to find the HCF of the two terms in the expression. The HCF of the two terms is the largest expression (biggest number of factors) you can find which fits into both terms without a remainder.

    The terms x and 5y do not have any factors in common: they do not share factors of x, nor do they share any number factors. In a case like this, the largest common factor is 1.

    The HCF for the terms x and 5y is 1.


    Submit your answer as:
  2. Is the expression prime or not?

    Answer:
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]
    "Prime" means that something cannot be broken into smaller factors. For example, 7 is a prime number because it cannot be broken into smaller factors. Similarly, an expression is prime if it cannot be broken in to smaller factors.
    STEP: <no title>
    [−1 point ⇒ 0 / 1 points left]

    The question asks us to decide if the expression x+5y is prime or not. "Prime" means that something cannot be broken down into smaller factors. For example, 10 is not prime because it can be broken down to 2×5; but 11 is prime because it cannot be broken into smaller factors: it can only be written as 1×11. The same idea also applies to expressions: an expression is prime if it cannot be broken into factors.

    How does the "prime" concept apply to the expression x+5y? From above we know that the HCF is 1: that means we cannot break up the expression more than this:

    x+5y1×(x+5y)

    Therefore, the expression in this question is prime.

    The correct choice from the list is: It is prime.


    Submit your answer as:

Prime expressions

Answer the two questions which follow about the expression 12x+20

  1. What is the HCF (highest common factor) of the terms in the expression?

    Answer:
    expression
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]
    The highest common factor (HCF) of two things is the largest number (or biggest expression) that fits into both things. For example, the HCF of 8y2 and 12y is 4y because it fits into both of the expressions.
    STEP: <no title>
    [−1 point ⇒ 0 / 1 points left]

    We have the expression 12x+20 and we need to find the HCF of the two terms in the expression. The HCF of the two terms is the largest expression (biggest number of factors) you can find which fits into both terms without a remainder.

    The terms in this expression, which are 12x and 20, have a number in common, but the variable is not common (it appears only in one of the terms).

    The HCF for the terms 12x and 20 is 4.


    Submit your answer as:
  2. Is the expression prime or not?

    Answer:
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]
    "Prime" means that something cannot be broken into smaller factors. For example, 7 is a prime number because it cannot be broken into smaller factors. Similarly, an expression is prime if it cannot be broken in to smaller factors.
    STEP: <no title>
    [−1 point ⇒ 0 / 1 points left]

    The question asks us to decide if the expression 12x+20 is prime or not. "Prime" means that something cannot be broken down into smaller factors. For example, 10 is not prime because it can be broken down to 2×5; but 11 is prime because it cannot be broken into smaller factors: it can only be written as 1×11. The same idea also applies to expressions: an expression is prime if it cannot be broken into factors.

    How does the "prime" concept apply to the expression 12x+20? From above we know that the HCF is 4: that means we can break up the expression into the factors 4 and (3x+5), like this:

    12x+204×(3x+5)

    Therefore, the expression in this question is not prime.

    The correct choice from the list is: It is not prime.


    Submit your answer as:

Prime expressions

Answer the two questions which follow about the expression 5x+4

  1. What is the HCF (highest common factor) of the terms in the expression?

    Answer:
    expression
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]
    The highest common factor (HCF) of two things is the largest number (or biggest expression) that fits into both things. For example, the HCF of 8y2 and 12y is 4y because it fits into both of the expressions.
    STEP: <no title>
    [−1 point ⇒ 0 / 1 points left]

    We have the expression 5x+4 and we need to find the HCF of the two terms in the expression. The HCF of the two terms is the largest expression (biggest number of factors) you can find which fits into both terms without a remainder.

    The terms 5x and 4 do not have any factors in common: they do not share factors of x, nor do they share any number factors. In a case like this, the largest common factor is 1.

    The HCF for the terms 5x and 4 is 1.


    Submit your answer as:
  2. Is the expression prime or not?

    Answer:
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]
    "Prime" means that something cannot be broken into smaller factors. For example, 7 is a prime number because it cannot be broken into smaller factors. Similarly, an expression is prime if it cannot be broken in to smaller factors.
    STEP: <no title>
    [−1 point ⇒ 0 / 1 points left]

    The question asks us to decide if the expression 5x+4 is prime or not. "Prime" means that something cannot be broken down into smaller factors. For example, 10 is not prime because it can be broken down to 2×5; but 11 is prime because it cannot be broken into smaller factors: it can only be written as 1×11. The same idea also applies to expressions: an expression is prime if it cannot be broken into factors.

    How does the "prime" concept apply to the expression 5x+4? From above we know that the HCF is 1: that means we cannot break up the expression more than this:

    5x+41×(5x+4)

    Therefore, the expression in this question is prime.

    The correct choice from the list is: It is prime.


    Submit your answer as:

Introduction to factorisation: multiple choice

For the expression 4x+4, which choice below is the correctly factorised expression?

A 4(x1)
B 4x
C 4(x+1)
Answer:

The correctly factorised expression is: .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Use the distributive law to expand the expressions in the table. The correct choice must be equal to 4x+4 when it is expanded.


STEP: Check the answer choices by expanding them
[−1 point ⇒ 0 / 1 points left]

This question is about factorisation. We have the expression 4x+4 and we need to select the choice in the table which is the correctly factorised version of that expression.

Factorisation is the inverse (the reverse) of the distributive law. Factorising means "undoing" the distributive law. In other words, factorisation is when you "un-expand" or "undistribute" an expression. The picture below shows the relationship between distributing and factorising. When we multiply into the brackets we are distributing; when we put a number back outside of the brackets, we are factorising.

We can use the relationship above to find the correct answer to this question! Just start by expanding the first choice in the table with the distributive law to see if it is equal to 4x+4:

Expand theexpression...4(x1)...to see if youget 4x+44x4

Shucks: it did not work out correctly. We needed the expression 4x+4 but instead we got 4x4. Move on to checking the second choice in the table: expand it using the distributive law. Once you find the expression which is equal to 4x+4 when you expand it, you have the correct answer.

In this question the correct answer is 4(x+1) because when we distribute the 4 into the brackets we get 4x+4.

The correct choice from the table is: C.


Submit your answer as:

Introduction to factorisation: multiple choice

For the expression 4x2+5x, which choice below is the correctly factorised expression?

A x(4x2+5)
B 5x(4x+1)
C x(4x+5)
Answer:

The correctly factorised expression is: .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Use the distributive law to expand the expressions in the table. The correct choice must be equal to 4x2+5x when it is expanded.


STEP: Check the answer choices by expanding them
[−1 point ⇒ 0 / 1 points left]

This question is about factorisation. We have the expression 4x2+5x and we need to select the choice in the table which is the correctly factorised version of that expression.

Factorisation is the inverse (the reverse) of the distributive law. Factorising means "undoing" the distributive law. In other words, factorisation is when you "un-expand" or "undistribute" an expression. The picture below shows the relationship between distributing and factorising. When we multiply into the brackets we are distributing; when we put a number back outside of the brackets, we are factorising.

We can use the relationship above to find the correct answer to this question! Just start by expanding the first choice in the table with the distributive law to see if it is equal to 4x2+5x:

Expand theexpression...x(4x2+5)...to see if youget 4x2+5x4x3+5x

Shucks: it did not work out correctly. We needed the expression 4x2+5x but instead we got 4x3+5x. Move on to checking the second choice in the table: expand it using the distributive law. Once you find the expression which is equal to 4x2+5x when you expand it, you have the correct answer.

In this question the correct answer is x(4x+5) because when we distribute the x into the brackets we get 4x2+5x.

The correct choice from the table is: C.


Submit your answer as:

Introduction to factorisation: multiple choice

For the expression 5x22x, which choice below is the correctly factorised expression?

A x(5x2)
B x(5x22)
C 2x(5x+1)
Answer:

The correctly factorised expression is: .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Use the distributive law to expand the expressions in the table. The correct choice must be equal to 5x22x when it is expanded.


STEP: Check the answer choices by expanding them
[−1 point ⇒ 0 / 1 points left]

This question is about factorisation. We have the expression 5x22x and we need to select the choice in the table which is the correctly factorised version of that expression.

Factorisation is the inverse (the reverse) of the distributive law. Factorising means "undoing" the distributive law. In other words, factorisation is when you "un-expand" or "undistribute" an expression. The picture below shows the relationship between distributing and factorising. When we multiply into the brackets we are distributing; when we put a number back outside of the brackets, we are factorising.

We can use the relationship above to find the correct answer to this question! Just start by expanding the first choice in the table with the distributive law to see if it is equal to 5x22x:

Expand theexpression...x(5x2)...to see if youget 5x22x5x22x

Fantastic! When we multiply out the first expression we find it is equal to 5x22x; that means it is the correctly factorised version of the expression.

The correct choice from the table is: A.


Submit your answer as:

Factorisation with trinomials

Factorise the expression given below.

INSTRUCTION: If it cannot be factorised, type "prime" in the answer box.
8x10y10z2
Answer:
polynomial
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]
You need to figure out the HCF of all three terms!
STEP: Determine the HCF (Highest Common Factor) of the terms in the expression
[−1 point ⇒ 2 / 3 points left]

You have the expression 8x10y10z2 and you must factorise it. Factorisation is the inverse of the distributive law, which means it is when you "un-expand" or "undistribute" an expression.

The first thing you need to do is find the HCF of the expression. However, the expression is a trinomial, so you must find the HCF of all three terms! The terms are 8x, 10y and 10z2. Remember that when all of the terms are negative, you should include the negative sign in the HCF. The HCF of the terms is 2.


STEP: Pull out the HCF, and then use the distributive law to check your answer
[−2 points ⇒ 0 / 3 points left]

Once you know the HCF, you can factorise the expression. Start by putting the HCF in front of a pair of brackets, like this: 8x10y10z2=2×(more stuff). If you take 2 out of each term in the trinomial you will get 2(4x+5y+5z2).

Remember that you should always check your answer when you factorise. To check the answer, distribute the HCF back into the brackets to make sure that everything agrees with the original expression. Remember, you must distribute the 2 to all three terms in the brackets.

2(4x+5y+5z2)8x10y10z2original expressionAll right! this is the

The correct answer is: 2(4x+5y+5z2).


Submit your answer as:

Factorisation with trinomials

Factorise the expression given below.

INSTRUCTION: If it cannot be factorised, type "prime" in the answer box.
20a3+8a28a
Answer:
polynomial
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]
You need to figure out the HCF of all three terms!
STEP: Determine the HCF (Highest Common Factor) of the terms in the expression
[−1 point ⇒ 2 / 3 points left]

You have the expression 20a3+8a28a and you must factorise it. Factorisation is the inverse of the distributive law, which means it is when you "un-expand" or "undistribute" an expression.

The first thing you need to do is find the HCF of the expression. However, the expression is a trinomial, so you must find the HCF of all three terms! The terms are 20a3, 8a2 and 8a. The HCF for these three terms is 4a.


STEP: Pull out the HCF, and then use the distributive law to check your answer
[−2 points ⇒ 0 / 3 points left]

Once you know the HCF, you can factorise the expression. Start by putting the HCF in front of a pair of brackets, like this: 20a3+8a28a=4a×(more stuff). If you take 4a out of each term in the trinomial you will get 4a(5a2+2a2).

Remember that you should always check your answer when you factorise. To check the answer, distribute the HCF back into the brackets to make sure that everything agrees with the original expression. Remember, you must distribute the 4a to all three terms in the brackets.

4a(5a2+2a2)20a3+8a28aoriginal expressionAll right! this is the

The correct answer is: 4a(5a2+2a2).


Submit your answer as:

Factorisation with trinomials

Factorise the expression given below.

INSTRUCTION: If it cannot be factorised, type "prime" in the answer box.
8a212a10
Answer:
polynomial
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]
You need to figure out the HCF of all three terms!
STEP: Determine the HCF (Highest Common Factor) of the terms in the expression
[−1 point ⇒ 2 / 3 points left]

You have the expression 8a212a10 and you must factorise it. Factorisation is the inverse of the distributive law, which means it is when you "un-expand" or "undistribute" an expression.

The first thing you need to do is find the HCF of the expression. However, the expression is a trinomial, so you must find the HCF of all three terms! The terms are 8a2, 12a and 10. The HCF for these three terms is 2.


STEP: Pull out the HCF, and then use the distributive law to check your answer
[−2 points ⇒ 0 / 3 points left]

Once you know the HCF, you can factorise the expression. Start by putting the HCF in front of a pair of brackets, like this: 8a212a10=2×(more stuff). If you take 2 out of each term in the trinomial you will get 2(4a26a5).

Remember that you should always check your answer when you factorise. To check the answer, distribute the HCF back into the brackets to make sure that everything agrees with the original expression. Remember, you must distribute the 2 to all three terms in the brackets.

2(4a26a5)8a212a10original expressionAll right! this is the

The correct answer is: 2(4a26a5).


Submit your answer as:

Factorising x2+bx+c when c>0

Factorise the following expression into two binomials:

x2+4x+4

Your answer should look something like this: (x + 1)(x + 7).

Answer:
polynomial
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]
Start your answer by writing out: (x+_)(x+_). Then you must find two numbers which make the factorisation work out correctly.
STEP: Set up the binomials based on the x2 term
[−2 points ⇒ 2 / 4 points left]

We have the expression x2+4x+4 and we must factorise it.

To factorise an expression means to write the expression as a product. For example, if you factorise the number 10, you would write 25... 2 and 5 are factors of 10. In the case of a trinomial like x2+4x+4 factorising usually leads to two binomials: we want two binomials which have a product equal to x2+4x+4, just like 2 and 5 have a product of 10.

To work out the binomial factors requires patience, because there is no single calculation which will give us the answer. Rather we must take the time to work out the answer by trial and error (guess & check).

The first step is to set up the answer based on the x2 term in the trinomial. It tells us that the binomials must be like this:

(x+_)(x+_)

Imagine starting FOIL with these two brackets: you will get x2 from the two x's, which is the first part of x2+4x+4.


STEP: Find the numbers which belong in the binomials
[−2 points ⇒ 0 / 4 points left]

So far we have this: x2+4x+4=(x+_)(x+_). Now we need to figure out which numbers belong in the empty spaces. Whatever numbers we try to put in the binomials, we must check the binomials with FOIL to make sure that everything works out correctly.

We should try numbers which have a product of 4. This is because of the "Lasts" term in FOIL. For example, let's check what happens if we try the numbers 2 and 2:

(x+2)(x+2)expand the productUse FOIL tox2+2x+2x+4x2+4x+4trinomial we want?Is this the

Nice! We got the trinomial we needed, so the binomials must be correct!

The correct answer is: (x+2)(x+2).


Submit your answer as:

Factorising x2+bx+c when c>0

Factorise the following expression into two binomials:

x2+4x+3

Your answer should look something like this: (x + 1)(x + 7).

Answer:
polynomial
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]
Start your answer by writing out: (x+_)(x+_). Then you must find two numbers which make the factorisation work out correctly.
STEP: Set up the binomials based on the x2 term
[−2 points ⇒ 2 / 4 points left]

We have the expression x2+4x+3 and we must factorise it.

To factorise an expression means to write the expression as a product. For example, if you factorise the number 10, you would write 25... 2 and 5 are factors of 10. In the case of a trinomial like x2+4x+3 factorising usually leads to two binomials: we want two binomials which have a product equal to x2+4x+3, just like 2 and 5 have a product of 10.

To work out the binomial factors requires patience, because there is no single calculation which will give us the answer. Rather we must take the time to work out the answer by trial and error (guess & check).

The first step is to set up the answer based on the x2 term in the trinomial. It tells us that the binomials must be like this:

(x+_)(x+_)

Imagine starting FOIL with these two brackets: you will get x2 from the two x's, which is the first part of x2+4x+3.


STEP: Find the numbers which belong in the binomials
[−2 points ⇒ 0 / 4 points left]

So far we have this: x2+4x+3=(x+_)(x+_). Now we need to figure out which numbers belong in the empty spaces. Whatever numbers we try to put in the binomials, we must check the binomials with FOIL to make sure that everything works out correctly.

We should try numbers which have a product of 3. This is because of the "Lasts" term in FOIL. For example, let's check what happens if we try the numbers 1 and 3:

(x+1)(x+3)expand the productUse FOIL tox2+3x+x+3x2+4x+3trinomial we want?Is this the

Nice! We got the trinomial we needed, so the binomials must be correct!

The correct answer is: (x+1)(x+3).


Submit your answer as:

Factorising x2+bx+c when c>0

Factorise the following expression into two binomials:

x2+9x+18

Your answer should look something like this: (x + 1)(x + 7).

Answer:
polynomial
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]
Start your answer by writing out: (x+_)(x+_). Then you must find two numbers which make the factorisation work out correctly.
STEP: Set up the binomials based on the x2 term
[−2 points ⇒ 2 / 4 points left]

We have the expression x2+9x+18 and we must factorise it.

To factorise an expression means to write the expression as a product. For example, if you factorise the number 10, you would write 25... 2 and 5 are factors of 10. In the case of a trinomial like x2+9x+18 factorising usually leads to two binomials: we want two binomials which have a product equal to x2+9x+18, just like 2 and 5 have a product of 10.

To work out the binomial factors requires patience, because there is no single calculation which will give us the answer. Rather we must take the time to work out the answer by trial and error (guess & check).

The first step is to set up the answer based on the x2 term in the trinomial. It tells us that the binomials must be like this:

(x+_)(x+_)

Imagine starting FOIL with these two brackets: you will get x2 from the two x's, which is the first part of x2+9x+18.


STEP: Find the numbers which belong in the binomials
[−2 points ⇒ 0 / 4 points left]

So far we have this: x2+9x+18=(x+_)(x+_). Now we need to figure out which numbers belong in the empty spaces. Whatever numbers we try to put in the binomials, we must check the binomials with FOIL to make sure that everything works out correctly.

We should try numbers which have a product of 18. This is because of the "Lasts" term in FOIL. For example, let's check what happens if we try the numbers 3 and 6:

(x+3)(x+6)expand the productUse FOIL tox2+6x+3x+18x2+9x+18trinomial we want?Is this the

Nice! We got the trinomial we needed, so the binomials must be correct!

The correct answer is: (x+3)(x+6).


Submit your answer as:

Factorising a trinomial

Factorise the following:

q2+4q32
Answer: q2+4q32=
polynomial
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Start by writing q2+4q32=(q+)(q+). Choose any two factors of 32 that can replace the 's such that the two expressions on either side of the '=' sign become equal.


STEP: Factorise q2+4q32
[−2 points ⇒ 0 / 2 points left]

Factorisation usually requires patience: it is not something you can do by following a formula. So give yourself time to find the answer if it does not come quickly.

The most useful thing to look at to start factorisation is the constant term; in this case it is equal to 32. That means that when you factorise you must find two numbers which have a product of ('product' means multiply) 32. The bad news is that those two numbers must also agree with the other terms in the question, which often makes it a bit complex.

In this case it works out like this:

q2+4q32=(q4)(q+8)
NOTE: It is very important when you factorise to check your answer; you can do that by expanding the binomials (using FOIL). If you check and your answer does not work, make some changes and try again!

Submit your answer as:

Factorising a trinomial

Factorise the following:

x2+3x4
Answer: x2+3x4=
polynomial
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Start by writing x2+3x4=(x+)(x+). Choose any two factors of 4 that can replace the 's such that the two expressions on either side of the '=' sign become equal.


STEP: Factorise x2+3x4
[−2 points ⇒ 0 / 2 points left]

Factorisation usually requires patience: it is not something you can do by following a formula. So give yourself time to find the answer if it does not come quickly.

The most useful thing to look at to start factorisation is the constant term; in this case it is equal to 4. That means that when you factorise you must find two numbers which have a product of ('product' means multiply) 4. The bad news is that those two numbers must also agree with the other terms in the question, which often makes it a bit complex.

In this case it works out like this:

x2+3x4=(x1)(x+4)
NOTE: It is very important when you factorise to check your answer; you can do that by expanding the binomials (using FOIL). If you check and your answer does not work, make some changes and try again!

Submit your answer as:

Factorising a trinomial

Factorise the following:

g211g+24
Answer: g211g+24=
polynomial
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Start by writing g211g+24=(g+)(g+). Choose any two factors of +24 that can replace the 's such that the two expressions on either side of the '=' sign become equal.


STEP: Factorise g211g+24
[−2 points ⇒ 0 / 2 points left]

Factorisation usually requires patience: it is not something you can do by following a formula. So give yourself time to find the answer if it does not come quickly.

The most useful thing to look at to start factorisation is the constant term; in this case it is equal to +24. That means that when you factorise you must find two numbers which have a product of ('product' means multiply) +24. The bad news is that those two numbers must also agree with the other terms in the question, which often makes it a bit complex.

In this case it works out like this:

g211g+24=(g8)(g3)
NOTE: It is very important when you factorise to check your answer; you can do that by expanding the binomials (using FOIL). If you check and your answer does not work, make some changes and try again!

Submit your answer as:

Factorising quadratic trinomials

Factorise the following expression if possible. If not possible, type "prime" in the answer box.

20z239z+18
Answer:
polynomial
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]
Here are two hints:
  1. use the signs in the trinomial to figure out what the signs in the binomials should be
  2. be careful to conclude that the trinomial is prime if you try a few combinations and they all fail: you need to try all possible combinations of numbers to get the quadratic and the constant term. If none of those combinations work correctly, then the trinomial is prime

STEP: <no title>
[−4 points ⇒ 0 / 4 points left]

This question is tough: both the quadratic coefficient and the constant term are composite numbers (not prime), which means there are a lot of combinations of numbers that you may need to try to get the answer. Ugh.

Use the signs in the trinomial as a guide for the signs in the binomials: since the constant term is positive but the linear term is negative, the signs in the binomials must both be negatives. Once you know what to expect with the signs, start working with the numbers to try to find the factors.

Based on the trinomial, here is one possible starting point:

(20z_)(z_)

If this option does not lead to the correct factors, you may have to come back to this starting point and make a change. It is possible that the trinomial is prime, but remember that you must try all possible combinations of numbers to get the quadratic and the constant term from the trinomial; if none of those combinations work correctly, then the trinomial is prime.

The correct factors are (4z3)(5z6), because when you expand the brackets it works out:

(4z3)(5z6)=20z224z15z+18=20z239z+18

The correct answer is: (4z3)(5z6).


Submit your answer as:

Factorising quadratic trinomials

Factorise the following expression if possible. If not possible, type "prime" in the answer box.

4a2+21a+27
Answer:
polynomial
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]
Here are two hints:
  1. use the signs in the trinomial to figure out what the signs in the binomials should be
  2. be careful to conclude that the trinomial is prime if you try a few combinations and they all fail: you need to try all possible combinations of numbers to get the quadratic and the constant term. If none of those combinations work correctly, then the trinomial is prime

STEP: <no title>
[−4 points ⇒ 0 / 4 points left]

This question is tough: both the quadratic coefficient and the constant term are composite numbers (not prime), which means there are a lot of combinations of numbers that you may need to try to get the answer. Ugh.

Use the signs in the trinomial as a guide for the signs in the binomials: since all of the terms in the trinomial are positive, the factors must both have plus signs. Once you know what to expect with the signs, start working with the numbers to try to find the factors.

Based on the trinomial, here is one possible starting point:

(2a+_)(2a+_)

If this option does not lead to the correct factors, you may have to come back to this starting point and make a change. It is possible that the trinomial is prime, but remember that you must try all possible combinations of numbers to get the quadratic and the constant term from the trinomial; if none of those combinations work correctly, then the trinomial is prime.

The correct factors are (4a+9)(a+3), because when you expand the brackets it works out:

(4a+9)(a+3)=4a2+12a+9a+27=4a2+21a+27

The correct answer is: (4a+9)(a+3).


Submit your answer as:

Factorising quadratic trinomials

Factorise the following expression if possible. If not possible, type "prime" in the answer box.

10y2y24
Answer:
polynomial
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]
Here are two hints:
  1. use the signs in the trinomial to figure out what the signs in the binomials should be
  2. be careful to conclude that the trinomial is prime if you try a few combinations and they all fail: you need to try all possible combinations of numbers to get the quadratic and the constant term. If none of those combinations work correctly, then the trinomial is prime

STEP: <no title>
[−4 points ⇒ 0 / 4 points left]

This question is tough: both the quadratic coefficient and the constant term are composite numbers (not prime), which means there are a lot of combinations of numbers that you may need to try to get the answer. Ugh.

Use the signs in the trinomial as a guide for the signs in the binomials: since the constant term is negative, the signs in the binomials must be opposites (one positive and one negative). Once you know what to expect with the signs, start working with the numbers to try to find the factors.

Based on the trinomial, here is one possible starting point:

(5y+_)(2y_)

If this option does not lead to the correct factors, you may have to come back to this starting point and make a change. It is possible that the trinomial is prime, but remember that you must try all possible combinations of numbers to get the quadratic and the constant term from the trinomial; if none of those combinations work correctly, then the trinomial is prime.

The correct factors are (5y8)(2y+3), because when you expand the brackets it works out:

(5y8)(2y+3)=10y2+15y16y24=10y2y24

The correct answer is: (5y8)(2y+3).


Submit your answer as:

Factorising quadratic expressions

Answer the following two questions about the expression x29x+20.

  1. Factorise the expression. If the expression cannot be factorised, type "prime" in the answer box.

    Answer:
    polynomial
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    STEP: Check to see if the expression is a special pattern
    [−1 point ⇒ 2 / 3 points left]

    The question gives the expression x29x+20 and you must factorise it.

    The first thing you should do is check to see if the expression is either a difference of two squares or the square of a binomial. A difference of two squares must have two terms made of squared values, with a subtraction between them. For a square of a binomial, the coefficient of the middle term must be twice as much as the square root of the constant term.

    The expression x29x+20 cannot be a difference of two squares because it is a trinomial (a difference of two squares must be a binomial). What about the square of a binomial? The expression does not follow that pattern either because the constant term is not a square number!

    The expression in this question is neither a difference of two squares nor the square of a binomial - it does not have one of the special patterns.


    STEP: Find the binomials using the signs in the expression
    [−2 points ⇒ 0 / 3 points left]

    Since there is no pattern to follow, you should use the signs in the trinomial to get started with the factorisation. In this case, the constant term is positive which means that the signs in the binomials must be the same. Furthermore, since the middle term is negative, the signs in the binomials must both be ""

    x29x+20=(x?_)(x?_)

    From this point you need to figure out what numbers belong in the binomials. Remember that these numbers must have a product of 20. Also remember that you should check your answer using FOIL to make sure that it is correct!

    The correct binomial factors are: (x4)(x5). You can also write the answer as (x5)(x4).


    Submit your answer as:
  2. Which of the following choices completes the statement correctly about the expression?

    Answer: The signs in the binomials must be .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]
    The correct answer is related to the rules for multiplication of positives and negatives.
    STEP: <no title>
    [−1 point ⇒ 0 / 1 points left]

    This question is about the relationship between the signs in the binomials and the signs in the expression x29x+20. These signs are related to each other by the rules for multiplying positives and negatives, and the signs "meet" (multiply each other) according to the connections in FOIL.

    In this question, the constant term is 20. Since this number comes from a multiplication during the FOIL calculations, the rules for positives and negatives with multiplication are at work. Therefore, the numbers that create the 20 must have the same type of signs (either positive-positive or negative-negative) because when you multiply numbers with the same signs, you get a positive answer, like 20 in the expression x29x+20.

    The correct ending to the sentence is that the binomials must have: the same signs because the constant term is positive.


    Submit your answer as:

Factorising quadratic expressions

Answer the following two questions about the expression y2+5y+6.

  1. Factorise the expression. If the expression cannot be factorised, type "prime" in the answer box.

    Answer:
    polynomial
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    STEP: Check to see if the expression is a special pattern
    [−1 point ⇒ 2 / 3 points left]

    The question gives the expression y2+5y+6 and you must factorise it.

    The first thing you should do is check to see if the expression is either a difference of two squares or the square of a binomial. A difference of two squares must have two terms made of squared values, with a subtraction between them. For a square of a binomial, the coefficient of the middle term must be twice as much as the square root of the constant term.

    The expression y2+5y+6 cannot be a difference of two squares because it is a trinomial (a difference of two squares must be a binomial). What about the square of a binomial? The expression does not follow that pattern either because the constant term is not a square number!

    The expression in this question is neither a difference of two squares nor the square of a binomial - it does not have one of the special patterns.


    STEP: Find the binomials using the signs in the expression
    [−2 points ⇒ 0 / 3 points left]

    Since there is no pattern to follow, you should use the signs in the trinomial to get started with the factorisation. In this case, the constant term is positive which means that the signs in the binomials must be the same. Furthermore, since the middle term is positive, the signs in the binomials must both be "+"

    y2+5y+6=(y+?_)(y+?_)

    From this point you need to figure out what numbers belong in the binomials. Remember that these numbers must have a product of 6. Also remember that you should check your answer using FOIL to make sure that it is correct!

    The correct binomial factors are: (y+3)(y+2). You can also write the answer as (y+2)(y+3).


    Submit your answer as:
  2. Which of the following choices completes the statement correctly about the expression?

    Answer: The signs in the binomials must be .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]
    The correct answer is related to the rules for multiplication of positives and negatives.
    STEP: <no title>
    [−1 point ⇒ 0 / 1 points left]

    This question is about the relationship between the signs in the binomials and the signs in the expression y2+5y+6. These signs are related to each other by the rules for multiplying positives and negatives, and the signs "meet" (multiply each other) according to the connections in FOIL.

    In this question, the constant term is 6. Since this number comes from a multiplication during the FOIL calculations, the rules for positives and negatives with multiplication are at work. Therefore, the numbers that create the 6 must have the same type of signs (either positive-positive or negative-negative) because when you multiply numbers with the same signs, you get a positive answer, like 6 in the expression y2+5y+6.

    The correct ending to the sentence is that the binomials must have: the same signs because the constant term is positive.


    Submit your answer as:

Factorising quadratic expressions

Answer the following two questions about the expression y264.

  1. Factorise the expression. If the expression cannot be factorised, type "prime" in the answer box.

    Answer:
    polynomial
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    STEP: Check to see if the expression is a special pattern
    [−1 point ⇒ 2 / 3 points left]

    The question gives the expression y264 and you must factorise it.

    The first thing you should do is check to see if the expression is either a difference of two squares or the square of a binomial. A difference of two squares must have two terms made of squared values, with a subtraction between them. For a square of a binomial, the coefficient of the middle term must be twice as much as the square root of the constant term.

    The binomial y264 is in fact a difference of two squares! That means that the binomials follow the pattern for a difference of two squares: they must be exactly the same except for the signs!

    The question gives you an expression with the difference of two squares pattern.


    STEP: Find the binomials using the pattern
    [−2 points ⇒ 0 / 3 points left]

    When you factorise a difference of two squares, you always get two binomials which are the same except for the signs: one binomial will have addition and the other must have subtraction.

    y264=(y+8)(y8)

    Remember that it is always a good idea to check your binomial factors with FOIL in order to make sure that everything works out: it is critical that you get it exactly right!

    The correct binomial factors are: (y+8)(y8). You can also write the answer as (y8)(y+8).


    Submit your answer as:
  2. Which of the following choices completes the statement correctly about the expression?

    Answer: The signs in the binomials must be .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]
    The correct answer is related to the rules for multiplication of positives and negatives.
    STEP: <no title>
    [−1 point ⇒ 0 / 1 points left]

    This question is about the relationship between the signs in the binomials and the signs in the expression y264. These signs are related to each other by the rules for multiplying positives and negatives, and the signs "meet" (multiply each other) according to the connections in FOIL.

    In this question, the constant term is 64. Since this number comes from a multiplication during the FOIL calculations, the rules for positives and negatives with multiplication are at work. Therefore, the numbers that create the 64 must have opposite signs because a positive multiplied by a negative gives a negative answer, like the 64 in the expression y264.

    The correct ending to the sentence is that the binomials must have: opposite signs because the constant term is negative.


    Submit your answer as:

Factorising x2+bx+c when c<0

Factorise the following expression into two binomials:

x2x12

Your answer should look something like this: (x + 1)(x - 5).

Answer:
polynomial
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]
Start your answer by writing out: (x+_)(x_). The signs in the binomials must be opposites - think about why! Then you must find two numbers which make the factorisation work out correctly.
STEP: Set up the binomials based on the x2 term
[−2 points ⇒ 2 / 4 points left]

You have the expression x2x12 and you must factorise it. To work out the binomial factors often requires patience: there is no single calculation which will give you the answer. Rather you must take the time to work out the answer by trial and error (guess & check).

For the expression x2x12, the first and last terms both give big hints about the binomials that you need. The first term suggests that the factors must both have x in them: (x)(x).

The last term in x2x12 tells you about the signs in the binomials. Remember that the number 12 usually comes from multiplying the numbers in the binomials (when you do FOIL). Therefore, the binomials must contain one negative sign and one positive sign. Therefore the binomials must be:

(x+_)(x_)

So at this point we have x2x12=(x+_)(x_).


STEP: Find the numbers which belong in the binomials
[−2 points ⇒ 0 / 4 points left]

Now you need to figure out which numbers belong in the empty spaces. Whatever numbers you try to put in the binomials, you must check the binomials with FOIL to make sure that everything works out correctly.

You should try numbers which have a product of 12. For example, let's check what happens with the numbers 1 and 12:

(x+1)(x12)expand the productUse FOIL tox212x+x12x211x12trinomial we want?Is this the

Nope! This is not the trinomial you needed: the x-term is not correct. (The other two terms are correct, but maths is very strict - you must get the trinomial x2x12 exactly, not almost the same trinomial.) That means that the binomials (x+1) and (x12) cannot be the correct factors.

Now stay patient: you need to try two other numbers which have a product of 12 and see if those numbers work out to make the trinomial x2x12. You must continue this process until you find two binomials which agree with the trinomial.

The correct answer is: (x4)(x+3). You can also write the answer as (x+3)(x4).


Submit your answer as:

Factorising x2+bx+c when c<0

Factorise the following expression into two binomials:

x2+x20

Your answer should look something like this: (x + 1)(x - 5).

Answer:
polynomial
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]
Start your answer by writing out: (x+_)(x_). The signs in the binomials must be opposites - think about why! Then you must find two numbers which make the factorisation work out correctly.
STEP: Set up the binomials based on the x2 term
[−2 points ⇒ 2 / 4 points left]

You have the expression x2+x20 and you must factorise it. To work out the binomial factors often requires patience: there is no single calculation which will give you the answer. Rather you must take the time to work out the answer by trial and error (guess & check).

For the expression x2+x20, the first and last terms both give big hints about the binomials that you need. The first term suggests that the factors must both have x in them: (x)(x).

The last term in x2+x20 tells you about the signs in the binomials. Remember that the number 20 usually comes from multiplying the numbers in the binomials (when you do FOIL). Therefore, the binomials must contain one negative sign and one positive sign. Therefore the binomials must be:

(x+_)(x_)

So at this point we have x2+x20=(x+_)(x_).


STEP: Find the numbers which belong in the binomials
[−2 points ⇒ 0 / 4 points left]

Now you need to figure out which numbers belong in the empty spaces. Whatever numbers you try to put in the binomials, you must check the binomials with FOIL to make sure that everything works out correctly.

You should try numbers which have a product of 20. For example, let's check what happens with the numbers 1 and 20:

(x+1)(x20)expand the productUse FOIL tox220x+x20x219x20trinomial we want?Is this the

Nope! This is not the trinomial you needed: the x-term is not correct. (The other two terms are correct, but maths is very strict - you must get the trinomial x2+x20 exactly, not almost the same trinomial.) That means that the binomials (x+1) and (x20) cannot be the correct factors.

Now stay patient: you need to try two other numbers which have a product of 20 and see if those numbers work out to make the trinomial x2+x20. You must continue this process until you find two binomials which agree with the trinomial.

The correct answer is: (x+5)(x4). You can also write the answer as (x4)(x+5).


Submit your answer as:

Factorising x2+bx+c when c<0

Factorise the following expression into two binomials:

x2x30

Your answer should look something like this: (x + 1)(x - 5).

Answer:
polynomial
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]
Start your answer by writing out: (x+_)(x_). The signs in the binomials must be opposites - think about why! Then you must find two numbers which make the factorisation work out correctly.
STEP: Set up the binomials based on the x2 term
[−2 points ⇒ 2 / 4 points left]

You have the expression x2x30 and you must factorise it. To work out the binomial factors often requires patience: there is no single calculation which will give you the answer. Rather you must take the time to work out the answer by trial and error (guess & check).

For the expression x2x30, the first and last terms both give big hints about the binomials that you need. The first term suggests that the factors must both have x in them: (x)(x).

The last term in x2x30 tells you about the signs in the binomials. Remember that the number 30 usually comes from multiplying the numbers in the binomials (when you do FOIL). Therefore, the binomials must contain one negative sign and one positive sign. Therefore the binomials must be:

(x+_)(x_)

So at this point we have x2x30=(x+_)(x_).


STEP: Find the numbers which belong in the binomials
[−2 points ⇒ 0 / 4 points left]

Now you need to figure out which numbers belong in the empty spaces. Whatever numbers you try to put in the binomials, you must check the binomials with FOIL to make sure that everything works out correctly.

You should try numbers which have a product of 30. For example, let's check what happens with the numbers 1 and 30:

(x+1)(x30)expand the productUse FOIL tox230x+x30x229x30trinomial we want?Is this the

Nope! This is not the trinomial you needed: the x-term is not correct. (The other two terms are correct, but maths is very strict - you must get the trinomial x2x30 exactly, not almost the same trinomial.) That means that the binomials (x+1) and (x30) cannot be the correct factors.

Now stay patient: you need to try two other numbers which have a product of 30 and see if those numbers work out to make the trinomial x2x30. You must continue this process until you find two binomials which agree with the trinomial.

The correct answer is: (x6)(x+5). You can also write the answer as (x+5)(x6).


Submit your answer as:

Factorisation: the distributive law backwards

Factorise the following expression by taking the HCF (Highest Common Factor) of the terms outside of brackets:

7x+7y

Your answer should look something like this: 3(x + 7).

Answer:
polynomial
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]
You need to figure out the HCF of the two terms first; then use the HCF to get the answer.
STEP: Determine the HCF (Highest Common Factor) of the terms in the expression
[−1 point ⇒ 2 / 3 points left]

We have the expression 7x+7y and we must factorise it. Factorisation means that we need to "undo" the distributive law - factorisation is the inverse of distributing. In other words, factorisation is when you "un-expand" or "undistribute" an expression.

The first thing we need to do is find the HCF of the two terms in the expression. (Remember that the HCF is the largest number or value which fits into both of the terms.) For the expression 7x+7y the HCF of the terms is 7.


STEP: Pull out the HCF, and then use the distributive law to check your answer
[−2 points ⇒ 0 / 3 points left]

Once you know the HCF, you can factorise the expression. As mentioned above, factorisation is when we "undo" the distributive law. The picture below shows the relationship between distributing and factorising. When you multiply into the brackets you are distributing; when you put a number back outside of the brackets, you are factorising.

In this question we are factorising, so start by putting the HCF in front of a pair of brackets, like this: 7x+7y=7(more stuff). Then you must fill in the brackets so that the expressions are equal.

In this case, if you take 7 out of the terms in the expression you will get 7(x+y). Can you see why factorisation means "undistributing"?

Checking the answer: When we factorise, we should always take the time necessary to check the answer. To check the answer we need to distribute the HCF, 7, back into the brackets to make sure that everything agrees with the original expression.

7(x+y)back into the bracketsMultiply the HCF7x+7yoriginal expressionCool - this is the

The correct answer is: 7(x+y).


Submit your answer as:

Factorisation: the distributive law backwards

Factorise the following expression by taking the HCF (Highest Common Factor) of the terms outside of brackets:

6x+6y

Your answer should look something like this: 3(x + 7).

Answer:
polynomial
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]
You need to figure out the HCF of the two terms first; then use the HCF to get the answer.
STEP: Determine the HCF (Highest Common Factor) of the terms in the expression
[−1 point ⇒ 2 / 3 points left]

We have the expression 6x+6y and we must factorise it. Factorisation means that we need to "undo" the distributive law - factorisation is the inverse of distributing. In other words, factorisation is when you "un-expand" or "undistribute" an expression.

The first thing we need to do is find the HCF of the two terms in the expression. (Remember that the HCF is the largest number or value which fits into both of the terms.) For the expression 6x+6y the HCF of the terms is 6.


STEP: Pull out the HCF, and then use the distributive law to check your answer
[−2 points ⇒ 0 / 3 points left]

Once you know the HCF, you can factorise the expression. As mentioned above, factorisation is when we "undo" the distributive law. The picture below shows the relationship between distributing and factorising. When you multiply into the brackets you are distributing; when you put a number back outside of the brackets, you are factorising.

In this question we are factorising, so start by putting the HCF in front of a pair of brackets, like this: 6x+6y=6(more stuff). Then you must fill in the brackets so that the expressions are equal.

In this case, if you take 6 out of the terms in the expression you will get 6(x+y). Can you see why factorisation means "undistributing"?

Checking the answer: When we factorise, we should always take the time necessary to check the answer. To check the answer we need to distribute the HCF, 6, back into the brackets to make sure that everything agrees with the original expression.

6(x+y)back into the bracketsMultiply the HCF6x+6yoriginal expressionCool - this is the

The correct answer is: 6(x+y).


Submit your answer as:

Factorisation: the distributive law backwards

Factorise the following expression by taking the HCF (Highest Common Factor) of the terms outside of brackets:

2x8

Your answer should look something like this: 3(x + 7).

Answer:
polynomial
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]
You need to figure out the HCF of the two terms first; then use the HCF to get the answer.
STEP: Determine the HCF (Highest Common Factor) of the terms in the expression
[−1 point ⇒ 2 / 3 points left]

We have the expression 2x8 and we must factorise it. Factorisation means that we need to "undo" the distributive law - factorisation is the inverse of distributing. In other words, factorisation is when you "un-expand" or "undistribute" an expression.

The first thing we need to do is find the HCF of the two terms in the expression. (Remember that the HCF is the largest number or value which fits into both of the terms.) For the expression 2x8 the HCF of the terms is 2.


STEP: Pull out the HCF, and then use the distributive law to check your answer
[−2 points ⇒ 0 / 3 points left]

Once you know the HCF, you can factorise the expression. As mentioned above, factorisation is when we "undo" the distributive law. The picture below shows the relationship between distributing and factorising. When you multiply into the brackets you are distributing; when you put a number back outside of the brackets, you are factorising.

In this question we are factorising, so start by putting the HCF in front of a pair of brackets, like this: 2x8=2(more stuff). Then you must fill in the brackets so that the expressions are equal.

In this case, if you take 2 out of the terms in the expression you will get 2(x4). Can you see why factorisation means "undistributing"?

Checking the answer: When we factorise, we should always take the time necessary to check the answer. To check the answer we need to distribute the HCF, 2, back into the brackets to make sure that everything agrees with the original expression.

2(x4)back into the bracketsMultiply the HCF2x8original expressionCool - this is the

The correct answer is: 2(x4).


Submit your answer as:

More complex HCFs

Factorise the following expression.

INSTRUCTION: If it cannot be factorised, type "prime" in the answer box.
8ac6bc
Answer:
polynomial
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]
You need to figure out the HCF of the two terms first; then use the HCF to get the answer.
STEP: Determine the HCF (Highest Common Factor) of the terms in the expression
[−1 point ⇒ 2 / 3 points left]

You have the expression 8ac6bc and you must factorise it. Factorisation is the inverse of the distributive law, which means it is when you "un-expand" or "undistribute" an expression.

The first thing you need to do is find the HCF of the two terms in the expression. The two terms are 8ac and 6bc. Remember that when all of the terms are negative, you should factorize the negative sign out of the expression in the HCF. The HCF of the terms is 2c.


STEP: Pull out the HCF, and then use the distributive law to check your answer
[−2 points ⇒ 0 / 3 points left]

Once you know the HCF, you can factorise the expression. Start by putting the HCF in front of a pair of brackets, like this: 8ac6bc=2c×(more stuff). If you take 2c out of the terms in the expression you will get 2c(4a+3b).

Remember that you should always check your answer when you factorise. To check the answer, distribute the HCF back into the brackets to make sure that everything agrees with the original expression.

2c(4a+3b)8ac6bcoriginal expressionBingo! this is the

The correct answer is: 2c(4a+3b).


Submit your answer as:

More complex HCFs

Factorise the following expression.

INSTRUCTION: If it cannot be factorised, type "prime" in the answer box.
24ab220b3
Answer:
polynomial
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]
You need to figure out the HCF of the two terms first; then use the HCF to get the answer.
STEP: Determine the HCF (Highest Common Factor) of the terms in the expression
[−1 point ⇒ 2 / 3 points left]

You have the expression 24ab220b3 and you must factorise it. Factorisation is the inverse of the distributive law, which means it is when you "un-expand" or "undistribute" an expression.

The first thing you need to do is find the HCF of the two terms in the expression. The two terms are 24ab2 and 20b3. The HCF of the terms is 4b2.


STEP: Pull out the HCF, and then use the distributive law to check your answer
[−2 points ⇒ 0 / 3 points left]

Once you know the HCF, you can factorise the expression. Start by putting the HCF in front of a pair of brackets, like this: 24ab220b3=4b2×(more stuff). If you take 4b2 out of the terms in the expression you will get 4b2(6a5b).

Remember that you should always check your answer when you factorise. To check the answer, distribute the HCF back into the brackets to make sure that everything agrees with the original expression.

4b2(6a5b)24ab220b3original expressionBingo! this is the

The correct answer is: 4b2(6a5b).


Submit your answer as:

More complex HCFs

Factorise the following expression.

INSTRUCTION: If it cannot be factorised, type "prime" in the answer box.
4x33yz
Answer:
string
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]
You need to figure out the HCF of the two terms first; then use the HCF to get the answer.
STEP: Determine the HCF (Highest Common Factor) of the terms in the expression
[−1 point ⇒ 2 / 3 points left]

You have the expression 4x33yz and you must factorise it. Factorisation is the inverse of the distributive law, which means it is when you "un-expand" or "undistribute" an expression.

The first thing you need to do is find the HCF of the two terms in the expression. The two terms are 4x3 and 3yz. These two terms have nothing in common at all, so the HCF is 1.


STEP: Pull out the HCF, and then use the distributive law to check your answer
[−2 points ⇒ 0 / 3 points left]

Once you know the HCF, you can factorise the expression. However, this is a special situation: the HCF is 1. That means that there are no common factors in the terms of the expression. Therefore, the expression must be prime!

A prime quantity has only two factors: 1 and itself. This expression has no other factors, only the HCF of 1 and the expression itself.

The correct answer is: prime.


Submit your answer as:

Factorising a difference of two squares

Factorise the following expression into two binomials.

4x249
Answer: 4x249=
polynomial
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]
The expression 4x249 has a special pattern: it is a difference of two squares. You can use this pattern to get to the answer quickly.
STEP: <no title>
[−2 points ⇒ 0 / 2 points left]

You need to factorise the expression 4x249. To work out the binomial factors often requires patience: however, there is a pattern in this expression which lets you get straight to the answer!

The expression 4x249 is a difference of two squares. (Both terms are squared values, and the subtraction is a difference.) The difference of two squares pattern is linked directly to binomials which are identical except for the signs in the binomials:

4x249=(2x+7)(2x7)
NOTE: You can check your answer by multiplying out:
(2x+7)(2x7)=x214x+14x49=4x249

The correct answer is: (2x+7)(2x7).


Submit your answer as:

Factorising a difference of two squares

Factorise the following expression into two binomials.

49x24
Answer: 49x24=
polynomial
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]
The expression 49x24 has a special pattern: it is a difference of two squares. You can use this pattern to get to the answer quickly.
STEP: <no title>
[−2 points ⇒ 0 / 2 points left]

You need to factorise the expression 49x24. To work out the binomial factors often requires patience: however, there is a pattern in this expression which lets you get straight to the answer!

The expression 49x24 is a difference of two squares. (Both terms are squared values, and the subtraction is a difference.) The difference of two squares pattern is linked directly to binomials which are identical except for the signs in the binomials:

49x24=(7x+2)(7x2)
NOTE: You can check your answer by multiplying out:
(7x+2)(7x2)=x214x+14x4=49x24

The correct answer is: (7x+2)(7x2).


Submit your answer as:

Factorising a difference of two squares

Factorise the following expression into two binomials.

4x281
Answer: 4x281=
polynomial
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]
The expression 4x281 has a special pattern: it is a difference of two squares. You can use this pattern to get to the answer quickly.
STEP: <no title>
[−2 points ⇒ 0 / 2 points left]

You need to factorise the expression 4x281. To work out the binomial factors often requires patience: however, there is a pattern in this expression which lets you get straight to the answer!

The expression 4x281 is a difference of two squares. (Both terms are squared values, and the subtraction is a difference.) The difference of two squares pattern is linked directly to binomials which are identical except for the signs in the binomials:

4x281=(2x+9)(2x9)
NOTE: You can check your answer by multiplying out:
(2x+9)(2x9)=x218x+18x81=4x281

The correct answer is: (2x+9)(2x9).


Submit your answer as:

Factorisation

Factorise the expression given below if possible. Type "prime" in the answer box if it cannot be factorised.

49y2+63yz
Answer:
polynomial
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]
You need to figure out the HCF of both terms in the expression!
STEP: <no title>
[−1 point ⇒ 2 / 3 points left]

You must factorise the expression 49y2+63yz. Start by figuring out the HCF of the expression.

The HCF for these terms is 7y.


STEP: <no title>
[−2 points ⇒ 0 / 3 points left]

Now factorise the HCF out of the expression. If you take 7y out of each term in the expression you will get 7y(7y+9z).

Finally, check your answer. Do this by distributing the HCF back into the brackets to make sure you get the original expression. Remember, you must distribute the 7y to each of the terms in the brackets.

7y(7y+9z) 49y2+63yz

The correct answer is: 7y(7y+9z).


Submit your answer as:

Factorisation

Factorise the expression given below if possible. Type "prime" in the answer box if it cannot be factorised.

4a2b24ab212ab
Answer:
polynomial
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]
You need to figure out the HCF of all three terms in the expression!
STEP: <no title>
[−1 point ⇒ 3 / 4 points left]

You must factorise the expression 4a2b24ab212ab. Start by figuring out the HCF of the expression. When dealing with a trinomial, the HCF must come from all three of the terms in the expression.

The HCF for these terms is 4ab.


STEP: <no title>
[−3 points ⇒ 0 / 4 points left]

Now factorise the HCF out of the expression. If you take 4ab out of each term in the expression you will get 4ab(a6b3).

Finally, check your answer. Do this by distributing the HCF back into the brackets to make sure you get the original expression. Remember, you must distribute the 4ab to each of the terms in the brackets.

4ab(a6b3) 4a2b24ab212ab

The correct answer is: 4ab(a6b3).


Submit your answer as:

Factorisation

Factorise the expression given below if possible. Type "prime" in the answer box if it cannot be factorised.

5x4+5x3+4
Answer:
string
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]
You need to figure out the HCF of all three terms in the expression!
STEP: <no title>
[−1 point ⇒ 3 / 4 points left]

You must factorise the expression 5x4+5x3+4. Start by figuring out the HCF of the expression. When dealing with a trinomial, the HCF must come from all three of the terms in the expression.

There are no factors which are shared in all of the terms; therefore, the HCF is 1.


STEP: <no title>
[−3 points ⇒ 0 / 4 points left]

At this point you want to factorise the expression. However, the HCF is 1 which means that there are no common factors in the terms of the expression. Therefore, 5x4+5x3+4 is a prime expression!

A prime expression (or number) has only two factors: 1 and itself. This expression has no other factors, only the HCF of 1 and the expression itself. Or to put it another way, the only way to "factorise" the expression is to write 5x4+5x3+4=1×(5x4+5x3+4).

The correct answer is: prime.


Submit your answer as:

Factorisation: common factors

Factorise: 6g(t6)+13(t6).

Answer:
polynomial
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Take out the common factor which is the expression in brackets.


STEP: Identify the common factor and factorise
[−2 points ⇒ 0 / 2 points left]

We take out the common bracket from the two terms to get:

6g(t6)+13(t6)=(t6)(6g+13)

The order of the brackets does not matter, since a×b is the same as b×a. Therefore the following is also correct:

6g(t6)+13(t6)=(6g+13)(t6)

Submit your answer as:

Factorisation: common factors

Factorise: 5f(r1)+6(r1).

Answer:
polynomial
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Take out the common factor which is the expression in brackets.


STEP: Identify the common factor and factorise
[−2 points ⇒ 0 / 2 points left]

We take out the common bracket from the two terms to get:

5f(r1)+6(r1)=(r1)(5f+6)

The order of the brackets does not matter, since a×b is the same as b×a. Therefore the following is also correct:

5f(r1)+6(r1)=(5f+6)(r1)

Submit your answer as:

Factorisation: common factors

Factorise the following: 5h(z8)+9(z8).

Answer:
polynomial
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Take out the common factor which is the expression in brackets.


STEP: Identify the common factor and factorise
[−2 points ⇒ 0 / 2 points left]

We take out the common bracket from the two terms to get:

5h(z8)+9(z8)=(z8)(5h+9)

The order of the brackets does not matter, since a×b is the same as b×a. Therefore the following is also correct:

5h(z8)+9(z8)=(5h+9)(z8)

Submit your answer as:

Factorising quadratic trinomials

Factorise the following expression if possible. If the expression cannot be factorised, write "prime" in the answer box.

2x2+x10
Answer: 2x2+x10=
polynomial
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]
Here are two things to keep in mind:
  1. Look at the constant term in the trinomial: it is negative, so the binomials must have opposite signs, like this: (_+_)(__)
  2. It is possible that the trinomial is prime; you need to try all possible combinations of numbers to get the quadratic and the constant term; if none of those combinations work correctly, then the trinomial is prime

STEP: Set up the binomials based on the 2x2 term and the signs
[−1 point ⇒ 2 / 3 points left]

You have the expression 2x2+x10 and you must factorise it. This question also has the challenging possibility that the expression is prime. To determine if the expression is prime, you must try all possible combinations of numbers to get the quadratic and the constant term from the trinomial; if none of those combinations work correctly, then the trinomial is prime.

For the factors of 2x2+x10, the following things must be true:

  • the 2x2 term means that the factors must start with 2x and x
  • since the constant term is negative, the signs in the binomials must be opposites (one positive and one negative)

From that information, here is an intelligent starting point:

(2x+_)(x_)

Remember that you may also need to try the option above with the signs swapped, (2x_)(x+_).


STEP: Find the numbers which belong in the binomials
[−2 points ⇒ 0 / 3 points left]

At this point you must try some numbers in the binomials; the numbers should have a product of 10. For example, see what happens if you try the numbers 10 and 1.

(2x+10)(x1)values 10 and 1Try out the2x22x+10x102x2+8x10we want or not?Is this the trinomial

Unfortunately, those binomials are not the correct factors. If there are any other possible combinations of numbers to put into the binomials, you must try them to see which combination works out. For example, you can try swapping the signs in the binomials: (2x10)(x+1) (which will lead to a different trinomial).

In the end, there is a way to factorise the trinomial, so it is not prime.

The correct answer is: (2x+5)(x2).


Submit your answer as:

Factorising quadratic trinomials

Factorise the following expression if possible. If the expression cannot be factorised, write "prime" in the answer box.

2x25x7
Answer: 2x25x7=
polynomial
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]
Here are two things to keep in mind:
  1. Look at the constant term in the trinomial: it is negative, so the binomials must have opposite signs, like this: (_+_)(__)
  2. It is possible that the trinomial is prime; you need to try all possible combinations of numbers to get the quadratic and the constant term; if none of those combinations work correctly, then the trinomial is prime

STEP: Set up the binomials based on the 2x2 term and the signs
[−1 point ⇒ 2 / 3 points left]

You have the expression 2x25x7 and you must factorise it. This question also has the challenging possibility that the expression is prime. To determine if the expression is prime, you must try all possible combinations of numbers to get the quadratic and the constant term from the trinomial; if none of those combinations work correctly, then the trinomial is prime.

For the factors of 2x25x7, the following things must be true:

  • the 2x2 term means that the factors must start with 2x and x
  • since the constant term is negative, the signs in the binomials must be opposites (one positive and one negative)

From that information, here is an intelligent starting point:

(2x+_)(x_)

Remember that you may also need to try the option above with the signs swapped, (2x_)(x+_).


STEP: Find the numbers which belong in the binomials
[−2 points ⇒ 0 / 3 points left]

At this point you must try some numbers in the binomials; the numbers should have a product of 7. For example, see what happens if you try the numbers 1 and 7.

(2x+1)(x7)values 1 and 7Try out the2x214x+x72x213x7we want or not?Is this the trinomial

Unfortunately, those binomials are not the correct factors. If there are any other possible combinations of numbers to put into the binomials, you must try them to see which combination works out. For example, you can try swapping the signs in the binomials: (2x1)(x+7) (which will lead to a different trinomial).

In the end, there is a way to factorise the trinomial, so it is not prime.

The correct answer is: (2x7)(x+1).


Submit your answer as:

Factorising quadratic trinomials

Factorise the following expression if possible. If the expression cannot be factorised, write "prime" in the answer box.

5x23x2
Answer: 5x23x2=
polynomial
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]
Here are two things to keep in mind:
  1. Look at the constant term in the trinomial: it is negative, so the binomials must have opposite signs, like this: (_+_)(__)
  2. It is possible that the trinomial is prime; you need to try all possible combinations of numbers to get the quadratic and the constant term; if none of those combinations work correctly, then the trinomial is prime

STEP: Set up the binomials based on the 5x2 term and the signs
[−1 point ⇒ 2 / 3 points left]

You have the expression 5x23x2 and you must factorise it. This question also has the challenging possibility that the expression is prime. To determine if the expression is prime, you must try all possible combinations of numbers to get the quadratic and the constant term from the trinomial; if none of those combinations work correctly, then the trinomial is prime.

For the factors of 5x23x2, the following things must be true:

  • the 5x2 term means that the factors must start with 5x and x
  • since the constant term is negative, the signs in the binomials must be opposites (one positive and one negative)

From that information, here is an intelligent starting point:

(5x+_)(x_)

Remember that you may also need to try the option above with the signs swapped, (5x_)(x+_).


STEP: Find the numbers which belong in the binomials
[−2 points ⇒ 0 / 3 points left]

At this point you must try some numbers in the binomials; the numbers should have a product of 2. For example, see what happens if you try the numbers 2 and 1.

(5x+2)(x1)values 2 and 1Try out the5x25x+2x25x23x2we want or not?Is this the trinomial

Great! We got the trinomial we needed, so the binomials must be correct.

The correct answer is: (5x+2)(x1).


Submit your answer as:

Factorisation: common factors

Factorise the following:

6fu+2uz
Answer:6fu+2uz=
polynomial
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Start by identifying the highest common factor (HCF) of both terms.


STEP: Identify the highest common factor (HCF) of both terms and factorise
[−2 points ⇒ 0 / 2 points left]

We take out the highest common factor (HCF) from both terms. In this case, the HCF is 2u.

6fu+2uz=2u(3f+z)

When you have factorised the HCF out of the expression, there cannot be any more common factors remaining in the brackets!


Submit your answer as:

Factorisation: common factors

Factorise:

21xp+33pa
Answer:21xp+33pa=
polynomial
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Start by identifying the highest common factor (HCF) of both terms.


STEP: Identify the highest common factor (HCF) of both terms and factorise
[−2 points ⇒ 0 / 2 points left]

We take out the highest common factor (HCF) from both terms. In this case, the HCF is 3p.

21xp+33pa=3p(7x+11a)

When you have factorised the HCF out of the expression, there cannot be any more common factors remaining in the brackets!


Submit your answer as:

Factorisation: common factors

Factorise:

42zs+6sg
Answer:42zs+6sg=
polynomial
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Start by identifying the highest common factor (HCF) of both terms.


STEP: Identify the highest common factor (HCF) of both terms and factorise
[−2 points ⇒ 0 / 2 points left]

We take out the highest common factor (HCF) from both terms. In this case, the HCF is 6s.

42zs+6sg=6s(7z+g)

When you have factorised the HCF out of the expression, there cannot be any more common factors remaining in the brackets!


Submit your answer as:

Factorising complex differences of two squares

Factorise the following expression into two binomials, if possible.

INSTRUCTION: If the expression cannot be factorised, type "prime" in the answer box.
16121h2
Answer:
polynomial
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]
What is the pattern in the binomials for a difference of two squares?
STEP: <no title>
[−2 points ⇒ 0 / 2 points left]

You need to factorise the expression 16121h2. Notice that all of the values in the expression are squared values (either square numbers or the square of a variable). Furthermore, the expression shows a difference (subtraction). All those facts together mean that the expression is a difference of two squares.

For the difference of two squares pattern, the binomials are always identical except for the signs in the binomials:

16121h2=(4+11h)(411h)

As always, you should check the binomials to make sure you get the exact expression you need.

(4+11h)(411h)expand the productUse FOIL to1644h+44h121h216121h2terms cancel each otherThe two middle

The correct answer is: (4+11h)(411h). The order of the binomials is not important, so you can also write the answer as (411h)(4+11h).


Submit your answer as:

Factorising complex differences of two squares

Factorise the following expression into two binomials, if possible.

INSTRUCTION: If the expression cannot be factorised, type "prime" in the answer box.
6449b2
Answer:
polynomial
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]
What is the pattern in the binomials for a difference of two squares?
STEP: <no title>
[−2 points ⇒ 0 / 2 points left]

You need to factorise the expression 6449b2. Notice that all of the values in the expression are squared values (either square numbers or the square of a variable). Furthermore, the expression shows a difference (subtraction). All those facts together mean that the expression is a difference of two squares.

For the difference of two squares pattern, the binomials are always identical except for the signs in the binomials:

6449b2=(8+7b)(87b)

As always, you should check the binomials to make sure you get the exact expression you need.

(8+7b)(87b)expand the productUse FOIL to6456b+56b49b26449b2terms cancel each otherThe two middle

The correct answer is: (8+7b)(87b). The order of the binomials is not important, so you can also write the answer as (87b)(8+7b).


Submit your answer as:

Factorising complex differences of two squares

Factorise the following expression into two binomials, if possible.

INSTRUCTION: If the expression cannot be factorised, type "prime" in the answer box.
25a2144
Answer:
polynomial
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]
What is the pattern in the binomials for a difference of two squares?
STEP: <no title>
[−2 points ⇒ 0 / 2 points left]

You need to factorise the expression 25a2144. Notice that all of the values in the expression are squared values (either square numbers or the square of a variable). Furthermore, the expression shows a difference (subtraction). All those facts together mean that the expression is a difference of two squares.

For the difference of two squares pattern, the binomials are always identical except for the signs in the binomials:

25a2144=(5a+12)(5a12)

As always, you should check the binomials to make sure you get the exact expression you need.

(5a+12)(5a12)expand the productUse FOIL to25a260a+60a14425a2144terms cancel each otherThe two middle

The correct answer is: (5a+12)(5a12). The order of the binomials is not important, so you can also write the answer as (5a12)(5a+12).


Submit your answer as:

Factorising x2+bx+c: true or false

The equation below shows a trinomial factorised into two binomial factors. Are the binomial factors correct or not? (Is the equation true or false?)

x210x+25=(x5)(x4)
Answer:
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]
Expand the binomials (with FOIL) to see if the you get the trinomial or not. Be careful to pay attention to the signs in the binomial.
STEP: <no title>
[−1 point ⇒ 0 / 1 points left]

Factorisation is the inverse of expanding something. When we factorise, we must break something apart into a product. For example, if you factorise the number 10, you write 25. For a trinomial, it is common that the factors are binomials, like what is shown in this question.

To see if the binomials are the correct factors, we must expand the product to find out if they are equal to the trinomial x210x+25 or not. That means using FOIL:

(x5)(x4)expand the productUse FOIL tox24x5x+20x29x+20trinomial we want?Is this the

We were supposed to get the trinomial x210x+25! However, the constant term and the x-term are both wrong. Therefore x210x+25 is not equal to (x5)(x4)

Notice that we checked the answer by expanding the binomials with FOIL and comparing the result to the trinomial in the question. Checking the answer with FOIL is a normal part of factorisation: you will do it over and over again when you are solving factorisation problems.

The correct choice is: False.


Submit your answer as:

Factorising x2+bx+c: true or false

The equation below shows a trinomial factorised into two binomial factors. Are the binomial factors correct or not? (Is the equation true or false?)

x24=(x2)(x2)
Answer:
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]
Expand the binomials (with FOIL) to see if the you get the trinomial or not. Be careful to pay attention to the signs in the binomial.
STEP: <no title>
[−1 point ⇒ 0 / 1 points left]

Factorisation is the inverse of expanding something. When we factorise, we must break something apart into a product. For example, if you factorise the number 10, you write 25. For a trinomial, it is common that the factors are binomials, like what is shown in this question.

To see if the binomials are the correct factors, we must expand the product to find out if they are equal to the trinomial x24 or not. That means using FOIL:

(x2)(x2)expand the productUse FOIL tox22x2x+4x24x+4trinomial we want?Is this the

We were supposed to get the trinomial x24! However, the constant term and the x-term are both wrong. Therefore x24 is not equal to (x2)(x2)

Notice that we checked the answer by expanding the binomials with FOIL and comparing the result to the trinomial in the question. Checking the answer with FOIL is a normal part of factorisation: you will do it over and over again when you are solving factorisation problems.

The correct choice is: False.


Submit your answer as:

Factorising x2+bx+c: true or false

The equation below shows a trinomial factorised into two binomial factors. Are the binomial factors correct or not? (Is the equation true or false?)

x2+10x+24=(x6)(x+4)
Answer:
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]
Expand the binomials (with FOIL) to see if the you get the trinomial or not. Be careful to pay attention to the signs in the binomial.
STEP: <no title>
[−1 point ⇒ 0 / 1 points left]

Factorisation is the inverse of expanding something. When we factorise, we must break something apart into a product. For example, if you factorise the number 10, you write 25. For a trinomial, it is common that the factors are binomials, like what is shown in this question.

To see if the binomials are the correct factors, we must expand the product to find out if they are equal to the trinomial x2+10x+24 or not. That means using FOIL:

(x6)(x+4)expand the productUse FOIL tox2+4x6x24x22x24trinomial we want?Is this the

These binomials are not the correct factors, because we did not get the trinomial in the question, x2+10x+24. The sign of the constant term is wrong, and the x-term is also wrong. Therefore, the equation x2+10x+24=(x+4)(x6) is not true!

Notice that we checked the answer by expanding the binomials with FOIL and comparing the result to the trinomial in the question. Checking the answer with FOIL is a normal part of factorisation: you will do it over and over again when you are solving factorisation problems.

The correct choice is: False.


Submit your answer as:

Factorising ax2+bx+c

Factorise the following expression into two binomials:

3x2+10x+3
Answer: 3x2+10x+3=
polynomial
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]
Since all of the terms in the trinomial are positive, the factors will be something like (_+_)(_+_).
STEP: Set up the binomials based on the 3x2 term and the signs
[−2 points ⇒ 2 / 4 points left]

You have the expression 3x2+10x+3 and you must factorise it. Factorising an expression usually requires patience, but in this case it is worse than usual because the quadratic coefficient is not 1: that adds to the complexity of the question.

Factorisation usually requires some trial and error, but you can usually use the quadratic term and the signs to start breaking down the factors. In this case:

  • the 3x2 term means that the factors must start with 3x and x
  • since all of the terms in the trinomial are positive, the factors must both have plus signs

That gives us:

(3x+_)(x+_)

(If you are not sure where these values came from, imagine starting FOIL with these two brackets and compare to the expression 3x2+10x+3.)


STEP: Find the numbers which belong in the binomials
[−2 points ⇒ 0 / 4 points left]

Now we need to figure out which numbers belong in the empty spaces. Whatever numbers we try to put in the binomials, they should have a product of 3 so that we can get the last term in 3x2+10x+3.

For example, let's check what happens if we try the numbers 3 and 1:

(3x+3)(x+1)values 3 and 1Try out the3x2+3x+3x+33x2+6x+3we want or not?Is this the trinomial

Shucks - we did not get the trinomial we needed because the x-term is not correct. That means that the binomials (3x+3) and (x+1) cannot be the correct factors.

You need to try another combination of numbers which have a product of 3. Note that swapping the 3 and 1 will give a different answer because of the 3x. In other words, even though (3x+3)(x+1) is not correct, (3x+1)(x+3) might be correct.

The correct factors are (3x+1)(x+3); you can test this using FOIL, and you will get 3x2+10x+3.

The correct answer is: (3x+1)(x+3).


Submit your answer as:

Factorising ax2+bx+c

Factorise the following expression into two binomials:

5x2+11x+2
Answer: 5x2+11x+2=
polynomial
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]
Since all of the terms in the trinomial are positive, the factors will be something like (_+_)(_+_).
STEP: Set up the binomials based on the 5x2 term and the signs
[−2 points ⇒ 2 / 4 points left]

You have the expression 5x2+11x+2 and you must factorise it. Factorising an expression usually requires patience, but in this case it is worse than usual because the quadratic coefficient is not 1: that adds to the complexity of the question.

Factorisation usually requires some trial and error, but you can usually use the quadratic term and the signs to start breaking down the factors. In this case:

  • the 5x2 term means that the factors must start with 5x and x
  • since all of the terms in the trinomial are positive, the factors must both have plus signs

That gives us:

(5x+_)(x+_)

(If you are not sure where these values came from, imagine starting FOIL with these two brackets and compare to the expression 5x2+11x+2.)


STEP: Find the numbers which belong in the binomials
[−2 points ⇒ 0 / 4 points left]

Now we need to figure out which numbers belong in the empty spaces. Whatever numbers we try to put in the binomials, they should have a product of 2 so that we can get the last term in 5x2+11x+2.

For example, let's check what happens if we try the numbers 2 and 1:

(5x+2)(x+1)values 2 and 1Try out the5x2+5x+2x+25x2+7x+2we want or not?Is this the trinomial

Shucks - we did not get the trinomial we needed because the x-term is not correct. That means that the binomials (5x+2) and (x+1) cannot be the correct factors.

You need to try another combination of numbers which have a product of 2. Note that swapping the 2 and 1 will give a different answer because of the 5x. In other words, even though (5x+2)(x+1) is not correct, (5x+1)(x+2) might be correct.

The correct factors are (5x+1)(x+2); you can test this using FOIL, and you will get 5x2+11x+2.

The correct answer is: (5x+1)(x+2).


Submit your answer as:

Factorising ax2+bx+c

Factorise the following expression into two binomials:

5x2+18x+9
Answer: 5x2+18x+9=
polynomial
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]
Since all of the terms in the trinomial are positive, the factors will be something like (_+_)(_+_).
STEP: Set up the binomials based on the 5x2 term and the signs
[−2 points ⇒ 2 / 4 points left]

You have the expression 5x2+18x+9 and you must factorise it. Factorising an expression usually requires patience, but in this case it is worse than usual because the quadratic coefficient is not 1: that adds to the complexity of the question.

Factorisation usually requires some trial and error, but you can usually use the quadratic term and the signs to start breaking down the factors. In this case:

  • the 5x2 term means that the factors must start with 5x and x
  • since all of the terms in the trinomial are positive, the factors must both have plus signs

That gives us:

(5x+_)(x+_)

(If you are not sure where these values came from, imagine starting FOIL with these two brackets and compare to the expression 5x2+18x+9.)


STEP: Find the numbers which belong in the binomials
[−2 points ⇒ 0 / 4 points left]

Now we need to figure out which numbers belong in the empty spaces. Whatever numbers we try to put in the binomials, they should have a product of 9 so that we can get the last term in 5x2+18x+9.

For example, let's check what happens if we try the numbers 3 and 3:

(5x+3)(x+3)values 3 and 3Try out the5x2+15x+3x+95x2+18x+9we want or not?Is this the trinomial

Nice! We got the trinomial we needed, so the binomials must be correct! It is quite nice that it worked out on the first try: it often takes more than one try.

The correct answer is: (5x+3)(x+3).


Submit your answer as:

Factorising quadratic trinomials: increasing complexity

Factorise the following expression. (It is possible to factorise the expression - it is not prime.)

15a2+13a6
Answer:
polynomial
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Here are two things to keep in mind: Look at the constant term in the trinomial: it is negative, so the binomials must have opposite signs, like this: (_+_)(__)


STEP: Set up the binomials based on the 15a2 term and the signs
[−2 points ⇒ 2 / 4 points left]

For this question you must factorise the expression 15a2+13a6. For the factors of 15a2+13a6, the following things must be true:

  • the coefficient of the quadratic term is a composite number, so there is more than one possibility for how to break it down in the binomial factors; the correct values might be 15a and a, or they could be 5a and 3a
  • since the constant term is negative, the signs in the binomials must be opposites (one positive and one negative)

From that information, here is an intelligent starting point.

(15a+_)(a_)

This is a good start, but keep in mind that it may prove to be wrong - it is not the only possibility; for example, the signs can be swapped, which would give a different answer.


STEP: Try some numbers in the binomials
[−2 points ⇒ 0 / 4 points left]

Now try some numbers in the binomials to get the constant term of 6. For example, see what happens if you try the numbers 1 and 6.

(5a+1)(3a6)values 1 and 6Try out the15a230a+3a615a227a6we want or not?Is this the trinomial

Unfortunately, those are not the correct factors. If there are any other possible combinations of numbers to put into the binomials, you must try them to see which combination works out. For example, you can try swapping the signs in the binomials: (5a1)(3a+6).

The correct answer is: (5a+6)(3a1).


Submit your answer as:

Factorising quadratic trinomials: increasing complexity

Factorise the following expression. (It is possible to factorise the expression - it is not prime.)

8a23a5
Answer:
polynomial
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Here are two things to keep in mind: Look at the constant term in the trinomial: it is negative, so the binomials must have opposite signs, like this: (_+_)(__)


STEP: Set up the binomials based on the 8a2 term and the signs
[−2 points ⇒ 2 / 4 points left]

For this question you must factorise the expression 8a23a5. For the factors of 8a23a5, the following things must be true:

  • the coefficient of the quadratic term is a composite number, so there is more than one possibility for how to break it down in the binomial factors; the correct values might be 8a and a, or they could be 4a and 2a
  • since the constant term is negative, the signs in the binomials must be opposites (one positive and one negative)

From that information, here is an intelligent starting point.

(8a+_)(a_)

This is a good start, but keep in mind that it may prove to be wrong - it is not the only possibility; for example, the signs can be swapped, which would give a different answer.


STEP: Try some numbers in the binomials
[−2 points ⇒ 0 / 4 points left]

Now try some numbers in the binomials to get the constant term of 5. For example, see what happens if you try the numbers 1 and 5.

(8a+1)(a5)values 1 and 5Try out the8a240a+a58a239a5we want or not?Is this the trinomial

Unfortunately, those are not the correct factors. If there are any other possible combinations of numbers to put into the binomials, you must try them to see which combination works out. For example, you can try swapping the signs in the binomials: (8a1)(a+5).

The correct answer is: (8a+5)(a1).


Submit your answer as:

Factorising quadratic trinomials: increasing complexity

Factorise the following expression. (It is possible to factorise the expression - it is not prime.)

6a217a10
Answer:
polynomial
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Here are two things to keep in mind: Look at the constant term in the trinomial: it is negative, so the binomials must have opposite signs, like this: (_+_)(__)


STEP: Set up the binomials based on the 6a2 term and the signs
[−2 points ⇒ 2 / 4 points left]

For this question you must factorise the expression 6a217a10. For the factors of 6a217a10, the following things must be true:

  • the coefficient of the quadratic term is a composite number, so there is more than one possibility for how to break it down in the binomial factors; the correct values might be 6a and a, or they could be 3a and 2a
  • since the constant term is negative, the signs in the binomials must be opposites (one positive and one negative)

From that information, here is an intelligent starting point.

(6a+_)(a_)

This is a good start, but keep in mind that it may prove to be wrong - it is not the only possibility; for example, the signs can be swapped, which would give a different answer.


STEP: Try some numbers in the binomials
[−2 points ⇒ 0 / 4 points left]

Now try some numbers in the binomials to get the constant term of 10. For example, see what happens if you try the numbers 1 and 10.

(3a+1)(2a10)values 1 and 10Try out the6a230a+2a106a228a10we want or not?Is this the trinomial

Unfortunately, those are not the correct factors. If there are any other possible combinations of numbers to put into the binomials, you must try them to see which combination works out. For example, you can try swapping the signs in the binomials: (3a1)(2a+10).

The correct answer is: (3a10)(2a+1).


Submit your answer as:

Factorising x2+bx+c: filling in the blank

The equation below shows an incomplete factorisation. What is the missing number?

x24x12=(x+?_)(x6)
Answer:
numeric
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]
Use FOIL on the binomials to see if they are equal. Be careful to pay attention to the signs in the binomial.
STEP: <no title>
[−1 point ⇒ 0 / 1 points left]

We have the equation x24x12=(x+?_)(x6), which shows an unfinished factorisation question. We must complete the factorisation by filling in the missing number.

To factorise an expression means to write the expression as a product. Factorising the trinomial x24x12 leads to two binomials: (x+?_)(x6). We need to put a number into the blank space so that the binomials have a product equal to x24x12.

We must think about FOIL when we factorise into binomials: if we expand the binomials using FOIL we must get the trinomial x24x12 (because the product must be equal to that trinomial).

(x+?_)(x6)x24x12

Imagine doing FOIL with these two brackets: the "Lasts" term must be equal to 12! What number should go into the blank space for that to work out?

Notice that the constant term in the trinomial is negative. That is why the signs in the binomials are opposites: negative multiplied by positive is negative! 2×6=12 .

The complete factorised expression is (x+2)(x6), and the answer is: 2.


Submit your answer as:

Factorising x2+bx+c: filling in the blank

The equation below shows an incomplete factorisation. What is the missing number?

x210x+21=(x?_)(x3)
Answer:
numeric
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]
Use FOIL on the binomials to see if they are equal. Be careful to pay attention to the signs in the binomial.
STEP: <no title>
[−1 point ⇒ 0 / 1 points left]

We have the equation x210x+21=(x?_)(x3), which shows an unfinished factorisation question. We must complete the factorisation by filling in the missing number.

To factorise an expression means to write the expression as a product. Factorising the trinomial x210x+21 leads to two binomials: (x?_)(x3). We need to put a number into the blank space so that the binomials have a product equal to x210x+21.

We must think about FOIL when we factorise into binomials: if we expand the binomials using FOIL we must get the trinomial x210x+21 (because the product must be equal to that trinomial).

(x?_)(x3)x210x+21

Imagine doing FOIL with these two brackets: the "Lasts" term must be equal to 21! What number should go into the blank space for that to work out?

Notice that the constant term in the trinomial is positive. That is why the signs in the binomials are the same (both negative): negative times negative is positive! 7×3=21 .

The complete factorised expression is (x7)(x3), and the answer is: 7.


Submit your answer as:

Factorising x2+bx+c: filling in the blank

The equation below shows an incomplete factorisation. What is the missing number?

x24x+3=(x?_)(x1)
Answer:
numeric
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]
Use FOIL on the binomials to see if they are equal. Be careful to pay attention to the signs in the binomial.
STEP: <no title>
[−1 point ⇒ 0 / 1 points left]

We have the equation x24x+3=(x?_)(x1), which shows an unfinished factorisation question. We must complete the factorisation by filling in the missing number.

To factorise an expression means to write the expression as a product. Factorising the trinomial x24x+3 leads to two binomials: (x?_)(x1). We need to put a number into the blank space so that the binomials have a product equal to x24x+3.

We must think about FOIL when we factorise into binomials: if we expand the binomials using FOIL we must get the trinomial x24x+3 (because the product must be equal to that trinomial).

(x?_)(x1)x24x+3

Imagine doing FOIL with these two brackets: the "Lasts" term must be equal to 3! What number should go into the blank space for that to work out?

Notice that the constant term in the trinomial is positive. That is why the signs in the binomials are the same (both negative): negative times negative is positive! 3×1=3 .

The complete factorised expression is (x3)(x1), and the answer is: 3.


Submit your answer as:

2. Solving quadratic equations

The difference of two squares

Solve for y if:

4y29=0

You can type your answers in any order.

Answer: y = or y
numeric
numeric
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

STEP: <no title>
[−2 points ⇒ 0 / 2 points left]

The first step in solving this problem is to factorise the binomial. This binomial is the difference of two squares. Once you have determined the two factors, apply the zero product rule by setting each factor equal to zero. You can then solve the two factors by isolating the variable, y.

4y29=0
(2y3)(2y+3)=0
2y3=02y+3=0y=32y=32

y=32 or y=32 .


Submit your answer as: and

Solving quadratic equations

Solve for c. You can type your answers in any order.

c2+12c+32=0
Answer: c= or c=
numeric
numeric
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

The method for solving quadratic equations is factorisation. After you factorise, you will use the two binomials you get from factorisation to find the solutions to the equation.


STEP: Factorise the left hand side
[−1 point ⇒ 1 / 2 points left]

Factorising the equation gives the following:

(c+8)(c+4)=0

STEP: Solve for c
[−1 point ⇒ 0 / 2 points left]

This means that one of these things must be true:

(c+8)=0or(c+4)=0

Therefore: c=8 or c=4.


Submit your answer as: and

Solving quadratic equations by inspection

Solve for a:

a2=4
Answer: a= or a=
fraction
fraction
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Which two numbers will give 4 when squared? Remember, when you square a negative number you get a positive number.


STEP: Solve for a
[−2 points ⇒ 0 / 2 points left]

We already know how to solve linear equations using inverse operations. But, this approach does not always work for non-linear equations. This equation has an a2 term, so it is not linear (it is quadratic).

When we solve equations, we are looking for the value of a that will make the left hand side equal to the right hand side. We will solve non-linear equations using inspection, which means looking carefully. This question is asking "What number (or numbers), when you square it, gives 4?".

Since 2×2=4, we know that 2 is a solution.

However, this is not the only solution. Remember, when you square a negative number you get a positive number.

(2)2=(2)(2)=4

Therefore, 2 is also a solution. Either of the two answers will make the equation true, so we write

a=2 or a=2
NOTE: Equations can have more than one solution - this equation has two answers. Both values make the equation true so we can't choose between them. Instead we give both answers.

Submit your answer as: and

Equations in standard form

Solve for a if:

(4a+1)(2a+3)=4a7

You may type your answers in any order.

Answer: a = or a =
numeric
numeric
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]
Standard form for a quadratic equation looks like this: 0=ax2+bx+c --OR-- ax2+bx+c=0. You just need to read the values of a, b and c from the equation.
STEP: <no title>
[−4 points ⇒ 0 / 4 points left]

The equation looks confusing because it is not in standard form. In "standard form," the values of a,b and c are as follows: 0=ax2+bx+c --or-- ax2+bx+c=0a is the coefficient of the quadratic term; b is the coefficient of the linear term; c is the value of the constant term.

  • You should first apply the distributive law to both sides of the equation.
  • Rearrange the equation into standard form by collecting all the terms on one side such that the value of a is positive and the equation is equal to zero.
  • Then the equation will look like a "normal" quadratic equation.
  • Find the factors and then solve for a.
(4a+1)(2a+3)=4a7(8a2+14a+3)=4a78a2+18a+10=0
4a+5=02a+2=0rule.Applying the zero producta=54a=1

Therefore, the roots of the equation are a=54 and a=1 .


Submit your answer as: and

Solving quadratic equations

Solve for x. You may type your answers in any order.

x23x42=2x
Answer: x= or x=
numeric
numeric
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Start by arranging the equation in standard form; then factorise!


STEP: <no title>
[−1 point ⇒ 3 / 4 points left]

You must first arrange the equation in standard form (the zero will be very important below!):

x23x42=2xx23x42+2x=0x2x42=0

STEP: <no title>
[−1 point ⇒ 2 / 4 points left]

Now factorise the trinomial on the left of the equation.

x2x42=0(x7)(x+6)=0

STEP: <no title>
[−2 points ⇒ 0 / 4 points left]

The equation above tells us that the binomials have a product of zero. That means that one of the binomials must be equal to zero:

If:(x7)(x+6)=0then: (x7)=0 or (x+6)=0x=6 or x=7

The solutions to the equation are x=6 or x=7.


Submit your answer as: and

Quadratic equations: Standard form

Solve for x if:

2x2+20x+46=x3

You may type your answers in any order.

Answer: x = or
x =
numeric
numeric
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Standard form for a quadratic equation looks like this:

0=ax2+bx+corax2+bx+c=0

You just need to read the values of a, b, and c from the equation.


STEP: Rearrange into standard form and factorise
[−4 points ⇒ 0 / 4 points left]

The equation might look confusing because it is not in standard form. In "standard form," the values of a, b, and c are as follows:

0=ax2+bx+corax2+bx+c=0

a is the coefficient of the quadratic term; b is the coefficient of the linear term; c is the value of the constant term.

You can rearrange the equation by collecting all the terms on one side which will result in a 0 on the other side of the equation. When you do this, make sure the quadratic coefficient is positive, and simplify the equation so that the powers of the terms are in descending order.

2x2+20x+46=x32x2+21x+49=0

The quadratic equation is now in "standard form" and ready to be factorised. Solve for x by making each factor equal to 0, and isolate the variable.

x+7=0x=72x+7=0x=72

The solutions for this equation are x=7 and x=72.


Submit your answer as: and

Solving quadratic equations

Solve for m. You can type your answers in any order.

0=4m2+24m20
Answer: m= or m=
numeric
numeric
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

You need to solve this equation by factorisation, but there is a step you can do before you factorise which will help make the problem easier!


STEP: <no title>
[−1 point ⇒ 3 / 4 points left]

All of the terms in this equation have a factor of 4. Divide the equation by 4 to remove this common factor and simplify the equation. Factorisation is easier with smaller numbers!

0=m26m+5

STEP: <no title>
[−1 point ⇒ 2 / 4 points left]

Factorise the right hand side:

0=(m5)(m1)

STEP: <no title>
[−2 points ⇒ 0 / 4 points left]

From the two binomials, you get two possibilities: either m1=0 or m5=0. From these two options, you get the answers: m=1 or m=5.


Submit your answer as: and

Equations with trinomials

Solve for b if:

b2+11b+24=0

You can type your answers in any order.

Answer: b = or b =
numeric
numeric
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

STEP: <no title>
[−2 points ⇒ 0 / 2 points left]

The first step in solving this problem is to factorise the trinomial. Once you have determined the two factors, apply the zero product rule by setting each factor equal to zero. You can then solve the two factors by isolating the variable, b.

b2+11b+24=0
(b+8)(b+3)=0
b+8=0b+3=0b=8b=3

b=8 or b=3.


Submit your answer as: and

Equations with trinomials

Solve for a if:

10a2+33a+20=0

You may type your answers in any order.

Answer: a = or a =
numeric
numeric
STEP: <no title>
[−2 points ⇒ 0 / 2 points left]

The first step in solving this problem is to factorise the trinomial.

10a2+33a+20=0(2a+5)(5a+4)=0

Once we have the two binomial factors, we use the zero product rule by setting each factor equal to zero. This leads to two separate equations. Then we can solve each of the equations by isolating the variable, a.

2a+5=05a+4=0a=52a=45

The correct answers are: a=52 or a=45.


Submit your answer as: and

Standard form for quadratic equations

Write this equation in standard form.

(3b2)(3b2)=19
INSTRUCTION: Write only the left side of the equation in the answer box.
Answer:

=0

polynomial
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Standard form for a quadratic equation is ax2+bx+c=0. You need to multiply the brackets and then simplify so that you can arrange the equation so that it matches standard form.


STEP: Expand the brackets and collect like terms
[−2 points ⇒ 1 / 3 points left]

We must expand the brackets. Also, remember to bring the constant over from the other side of the equation because standard form for a quadratic equation requires a zero on one side of the equation:

(3b2)(3b2)=199b26b6b+4+19=0

STEP: Simplify the equation
[−1 point ⇒ 0 / 3 points left]

Now simplify the expression on the left of the equation by collecting like terms.

9b26b6b+4+19=09b212b+23=0

The equation now matches the arrangement for standard form:ab2+bb+c=0; so we are done!

The final answer is 9b212b+23=0


Submit your answer as:

Equations with a difference of two squares

Solve for b if:

b249=0

You can type your answers in any order.

Answer: b = or b =
numeric
numeric
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

STEP: <no title>
[−2 points ⇒ 0 / 2 points left]

The first step in solving this problem is to factorise the binomial. This binomial is the the difference of two squares. When you have the two factors you set them equal to zero, applying the zero product rule. You then solve the two factors by isolating the variable, b.

b249=0
(b+7)(b7)=0
b+7=0b7=0b=7b=7

b=7 or b=7


Submit your answer as: and

Equations with trinomials

Solve for a if:

a2+13a+36=0

You can type your answers in any order.

Answer: a = or a =
numeric
numeric
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

STEP: <no title>
[−2 points ⇒ 0 / 2 points left]

The first step in solving this problem is to factorise the trinomial. When you have the two factors you set them equal to zero, applying the zero product rule. You then solve the two factors by isolating the variable, a.

a2+13a+36=0
(a+9)(a+4)=0
a+9=0a+4=0a=9a=4

a=9 or a=4.


Submit your answer as: and

Equations with trinomials

Solve for b if:

5b2+47b+18=0

You can type your answers in any order.

Answer: b = or b =
numeric
numeric
STEP: <no title>
[−2 points ⇒ 0 / 2 points left]

The first step in solving this problem is to factorise the trinomial. Once we have determined the two factors, we use the zero product rule by setting each factor equal to zero. We can then solve the two factors by isolating the variable, b.

5b2+47b+18=0(5b+2)(b+9)=0
5b+2=0b+9=0b=25b=9

The solutions to the equation are b=9 and b=25.


Submit your answer as: and

Solve quadratic equations by inspection

Solve for a:

(a+5)2=36
Answer: a= or a=
fraction
fraction
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Which two numbers will give 36 when squared? Remember, when you multiply a negative number by itself you get a positive number.


STEP: Determine two numbers that give 36 when squared
[−2 points ⇒ 1 / 3 points left]

We already know how to solve linear equations using inverse operations. But, this approach does not always work for non-linear equations. This equation has a a2 term, so it is not linear (it is quadratic). We use inspection to solve non-linear equations.

We will start by solving this equation for the whole bracket:

(a+5)2=36

To answer this question we ask "What number (or numbers), when you square it, gives you 36?".

Since 6×6=36, we know that 6 is a value for a+5.

However, this is not the only number that gives 36 when multiplied by itself. Remember, when you square a negative number you get a positive number:

(6)2=(6)(6)=36

Therefore, a+5 must be equal to 6 or 6.


STEP: Determine a
[−1 point ⇒ 0 / 3 points left]

Mathematically we write this as:

(a+5)2=36a+5=6 or a+5=6

Now we have created two linear equations that can each be solved separately for a, using inverse operations.

a+5=6a+55=65a=1

or

a+5=6a+55=65a=11

We can check our answers by substituting in the values for a:

  • If a=1: (a+5)2 = (1+5)2 = (6)2 = 36
  • If a=11: (a+5)2 = (11+5)2 = (6)2 = 36

Since these a-values make the equation true, we know that they are both valid answers.


Submit your answer as: and

Exercises

The difference of two squares

Solve for z if:

z21=0

You can type your answers in any order.

Answer: z = or z
numeric
numeric
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

STEP: <no title>
[−2 points ⇒ 0 / 2 points left]

The first step in solving this problem is to factorise the binomial. This binomial is the difference of two squares. Once you have determined the two factors, apply the zero product rule by setting each factor equal to zero. You can then solve the two factors by isolating the variable, z.

z21=0
(z+1)(z1)=0
z+1=0z1=0z=1z=1

z=1 or z=1 .


Submit your answer as: and

The difference of two squares

Solve for x if:

25x24=0

You can type your answers in any order.

Answer: x = or x
numeric
numeric
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

STEP: <no title>
[−2 points ⇒ 0 / 2 points left]

The first step in solving this problem is to factorise the binomial. This binomial is the difference of two squares. Once you have determined the two factors, apply the zero product rule by setting each factor equal to zero. You can then solve the two factors by isolating the variable, x.

25x24=0
(5x+2)(5x2)=0
5x+2=05x2=0x=25x=25

x=25 or x=25 .


Submit your answer as: and

The difference of two squares

Solve for y if:

49y24=0

You can type your answers in any order.

Answer: y = or y
numeric
numeric
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

STEP: <no title>
[−2 points ⇒ 0 / 2 points left]

The first step in solving this problem is to factorise the binomial. This binomial is the difference of two squares. Once you have determined the two factors, apply the zero product rule by setting each factor equal to zero. You can then solve the two factors by isolating the variable, y.

49y24=0
(7y+2)(7y2)=0
7y+2=07y2=0y=27y=27

y=27 or y=27 .


Submit your answer as: and

Solving quadratic equations

Solve for m. You can type your answers in any order.

m2+2m63=0
Answer: m= or m=
numeric
numeric
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

The method for solving quadratic equations is factorisation. After you factorise, you will use the two binomials you get from factorisation to find the solutions to the equation.


STEP: Factorise the left hand side
[−1 point ⇒ 1 / 2 points left]

Factorising the equation gives the following:

(m7)(m+9)=0

STEP: Solve for m
[−1 point ⇒ 0 / 2 points left]

This means that one of these things must be true:

(m7)=0or(m+9)=0

Therefore: m=9 or m=7.


Submit your answer as: and

Solving quadratic equations

Solve for y. You can type your answers in any order.

y211y+30=0
Answer: y= or y=
numeric
numeric
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

The method for solving quadratic equations is factorisation. After you factorise, you will use the two binomials you get from factorisation to find the solutions to the equation.


STEP: Factorise the left hand side
[−1 point ⇒ 1 / 2 points left]

Factorising the equation gives the following:

(y6)(y5)=0

STEP: Solve for y
[−1 point ⇒ 0 / 2 points left]

This means that one of these things must be true:

(y6)=0or(y5)=0

Therefore: y=5 or y=6.


Submit your answer as: and

Solving quadratic equations

Solve for x. You can type your answers in any order.

x2+4x45=0
Answer: x= or x=
numeric
numeric
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

The method for solving quadratic equations is factorisation. After you factorise, you will use the two binomials you get from factorisation to find the solutions to the equation.


STEP: Factorise the left hand side
[−1 point ⇒ 1 / 2 points left]

Factorising the equation gives the following:

(x5)(x+9)=0

STEP: Solve for x
[−1 point ⇒ 0 / 2 points left]

This means that one of these things must be true:

(x5)=0or(x+9)=0

Therefore: x=9 or x=5.


Submit your answer as: and

Solving quadratic equations by inspection

Solve for q:

4=q2
Answer: q= or q=
fraction
fraction
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Which two numbers will give 4 when squared? Remember, when you square a negative number you get a positive number.


STEP: Solve for q
[−2 points ⇒ 0 / 2 points left]

We already know how to solve linear equations using inverse operations. But, this approach does not always work for non-linear equations. This equation has a q2 term, so it is not linear (it is quadratic).

When we solve equations, we are looking for the value of q that will make the left hand side equal to the right hand side. We will solve non-linear equations using inspection, which means looking carefully. This question is asking "What number (or numbers), when you square it, gives 4?".

Since 2×2=4, we know that 2 is a solution.

However, this is not the only solution. Remember, when you square a negative number you get a positive number.

(2)2=(2)(2)=4

Therefore, 2 is also a solution. Either of the two answers will make the equation true, so we write

q=2 or q=2
NOTE: Equations can have more than one solution - this equation has two answers. Both values make the equation true so we can't choose between them. Instead we give both answers.

Submit your answer as: and

Solving quadratic equations by inspection

Solve for q:

16=q2
Answer: q= or q=
fraction
fraction
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Which two numbers will give 16 when squared? Remember, when you square a negative number you get a positive number.


STEP: Solve for q
[−2 points ⇒ 0 / 2 points left]

We already know how to solve linear equations using inverse operations. But, this approach does not always work for non-linear equations. This equation has a q2 term, so it is not linear (it is quadratic).

When we solve equations, we are looking for the value of q that will make the left hand side equal to the right hand side. We will solve non-linear equations using inspection, which means looking carefully. This question is asking "What number (or numbers), when you square it, gives 16?".

Since 4×4=16, we know that 4 is a solution.

However, this is not the only solution. Remember, when you square a negative number you get a positive number.

(4)2=(4)(4)=16

Therefore, 4 is also a solution. Either of the two answers will make the equation true, so we write

q=4 or q=4
NOTE: Equations can have more than one solution - this equation has two answers. Both values make the equation true so we can't choose between them. Instead we give both answers.

Submit your answer as: and

Solving quadratic equations by inspection

Solve for y:

64=y2
Answer: y= or y=
fraction
fraction
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Which two numbers will give 64 when squared? Remember, when you square a negative number you get a positive number.


STEP: Solve for y
[−2 points ⇒ 0 / 2 points left]

We already know how to solve linear equations using inverse operations. But, this approach does not always work for non-linear equations. This equation has a y2 term, so it is not linear (it is quadratic).

When we solve equations, we are looking for the value of y that will make the left hand side equal to the right hand side. We will solve non-linear equations using inspection, which means looking carefully. This question is asking "What number (or numbers), when you square it, gives 64?".

Since 8×8=64, we know that 8 is a solution.

However, this is not the only solution. Remember, when you square a negative number you get a positive number.

(8)2=(8)(8)=64

Therefore, 8 is also a solution. Either of the two answers will make the equation true, so we write

y=8 or y=8
NOTE: Equations can have more than one solution - this equation has two answers. Both values make the equation true so we can't choose between them. Instead we give both answers.

Submit your answer as: and

Equations in standard form

Solve for b if:

(5b+3)(b+2)=6(b+2)

You may type your answers in any order.

Answer: b = or b =
numeric
numeric
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]
Standard form for a quadratic equation looks like this: 0=ax2+bx+c --OR-- ax2+bx+c=0. You just need to read the values of a, b and c from the equation.
STEP: <no title>
[−4 points ⇒ 0 / 4 points left]

The equation looks confusing because it is not in standard form. In "standard form," the values of a,b and c are as follows: 0=ax2+bx+c --or-- ax2+bx+c=0a is the coefficient of the quadratic term; b is the coefficient of the linear term; c is the value of the constant term.

  • You should first apply the distributive law to both sides of the equation.
  • Rearrange the equation into standard form by collecting all the terms on one side such that the value of a is positive and the equation is equal to zero.
  • Then the equation will look like a "normal" quadratic equation.
  • Find the factors and then solve for b.
(5b+3)(b+2)=6(b+2)(5b2+13b+6)=6b125b2+19b+18=0
5b+9=0b+2=0rule.Applying the zero productb=95b=2

Therefore, the roots of the equation are b=2 and b=95 .


Submit your answer as: and

Equations in standard form

Solve for z if:

(3z+7)(z+8)=19z+47

You may type your answers in any order.

Answer: z = or z =
numeric
numeric
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]
Standard form for a quadratic equation looks like this: 0=ax2+bx+c --OR-- ax2+bx+c=0. You just need to read the values of a, b and c from the equation.
STEP: <no title>
[−4 points ⇒ 0 / 4 points left]

The equation looks confusing because it is not in standard form. In "standard form," the values of a,b and c are as follows: 0=ax2+bx+c --or-- ax2+bx+c=0a is the coefficient of the quadratic term; b is the coefficient of the linear term; c is the value of the constant term.

  • You should first apply the distributive law to both sides of the equation.
  • Rearrange the equation into standard form by collecting all the terms on one side such that the value of a is positive and the equation is equal to zero.
  • Then the equation will look like a "normal" quadratic equation.
  • Find the factors and then solve for z.
(3z+7)(z+8)=19z+47(3z2+31z+56)=19z+473z2+12z+9=0
3z+9=0z+1=0rule.Applying the zero productz=3z=1

Therefore, the roots of the equation are z=3 and z=1 .


Submit your answer as: and

Equations in standard form

Solve for z if:

(z+8)(z+8)=4(z+7)

You may type your answers in any order.

Answer: z = or z =
numeric
numeric
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]
Standard form for a quadratic equation looks like this: 0=ax2+bx+c --OR-- ax2+bx+c=0. You just need to read the values of a, b and c from the equation.
STEP: <no title>
[−4 points ⇒ 0 / 4 points left]

The equation looks confusing because it is not in standard form. In "standard form," the values of a,b and c are as follows: 0=ax2+bx+c --or-- ax2+bx+c=0a is the coefficient of the quadratic term; b is the coefficient of the linear term; c is the value of the constant term.

  • You should first apply the distributive law to both sides of the equation.
  • Rearrange the equation into standard form by collecting all the terms on one side such that the value of a is positive and the equation is equal to zero.
  • Then the equation will look like a "normal" quadratic equation.
  • Find the factors and then solve for z.
(z+8)(z+8)=4(z+7)(z2+16z+64)=4z+28z2+12z+36=0
z+6=0z+6=0rule.Applying the zero productz=6z=6

Therefore, the roots of the equation are z=6 and z=6 . This is a special case where both roots have the same solution, z=6.


Submit your answer as: and

Solving quadratic equations

Solve for p. You may type your answers in any order.

p2+p+2=2p
Answer: p= or p=
numeric
numeric
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Start by arranging the equation in standard form; then factorise!


STEP: <no title>
[−1 point ⇒ 3 / 4 points left]

You must first arrange the equation in standard form (the zero will be very important below!):

p2+p+2=2pp2+p+2+2p=0p2+3p+2=0

STEP: <no title>
[−1 point ⇒ 2 / 4 points left]

Now factorise the trinomial on the left of the equation.

p2+3p+2=0(p+1)(p+2)=0

STEP: <no title>
[−2 points ⇒ 0 / 4 points left]

The equation above tells us that the binomials have a product of zero. That means that one of the binomials must be equal to zero:

If:(p+1)(p+2)=0then: (p+1)=0 or (p+2)=0p=2 or p=1

The solutions to the equation are p=2 or p=1.


Submit your answer as: and

Solving quadratic equations

Solve for a. You may type your answers in any order.

a26a+15=2a
Answer: a= or a=
numeric
numeric
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Start by arranging the equation in standard form; then factorise!


STEP: <no title>
[−1 point ⇒ 3 / 4 points left]

You must first arrange the equation in standard form (the zero will be very important below!):

a26a+15=2aa26a+152a=0a28a+15=0

STEP: <no title>
[−1 point ⇒ 2 / 4 points left]

Now factorise the trinomial on the left of the equation.

a28a+15=0(a5)(a3)=0

STEP: <no title>
[−2 points ⇒ 0 / 4 points left]

The equation above tells us that the binomials have a product of zero. That means that one of the binomials must be equal to zero:

If:(a5)(a3)=0then: (a5)=0 or (a3)=0a=3 or a=5

The solutions to the equation are a=3 or a=5.


Submit your answer as: and

Solving quadratic equations

Solve for y. You may type your answers in any order.

y2+9y+13=5
Answer: y= or y=
numeric
numeric
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Start by arranging the equation in standard form; then factorise!


STEP: <no title>
[−1 point ⇒ 3 / 4 points left]

You must first arrange the equation in standard form (the zero will be very important below!):

y2+9y+13=5y2+9y+13+5=0y2+9y+18=0

STEP: <no title>
[−1 point ⇒ 2 / 4 points left]

Now factorise the trinomial on the left of the equation.

y2+9y+18=0(y+3)(y+6)=0

STEP: <no title>
[−2 points ⇒ 0 / 4 points left]

The equation above tells us that the binomials have a product of zero. That means that one of the binomials must be equal to zero:

If:(y+3)(y+6)=0then: (y+3)=0 or (y+6)=0y=6 or y=3

The solutions to the equation are y=6 or y=3.


Submit your answer as: and

Quadratic equations: Standard form

Solve for b if:

b2+19b+27=b2+3

You may type your answers in any order.

Answer: b = or
b =
numeric
numeric
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Standard form for a quadratic equation looks like this:

0=ax2+bx+corax2+bx+c=0

You just need to read the values of a, b, and c from the equation.


STEP: Rearrange into standard form and factorise
[−4 points ⇒ 0 / 4 points left]

The equation might look confusing because it is not in standard form. In "standard form," the values of a, b, and c are as follows:

0=ax2+bx+corax2+bx+c=0

a is the coefficient of the quadratic term; b is the coefficient of the linear term; c is the value of the constant term.

You can rearrange the equation by collecting all the terms on one side which will result in a 0 on the other side of the equation. When you do this, make sure the quadratic coefficient is positive, and simplify the equation so that the powers of the terms are in descending order.

b2+19b+27=b2+32b2+19b+24=0

The quadratic equation is now in "standard form" and ready to be factorised. Solve for b by making each factor equal to 0, and isolate the variable.

2b+3=0b=32b+8=0b=8

The solutions for this equation are b=8 and b=32.


Submit your answer as: and

Quadratic equations: Standard form

Solve for b if:

3b2+14b+40=2b2

You may type your answers in any order.

Answer: b = or
b =
numeric
numeric
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Standard form for a quadratic equation looks like this:

0=ax2+bx+corax2+bx+c=0

You just need to read the values of a, b, and c from the equation.


STEP: Rearrange into standard form and factorise
[−4 points ⇒ 0 / 4 points left]

The equation might look confusing because it is not in standard form. In "standard form," the values of a, b, and c are as follows:

0=ax2+bx+corax2+bx+c=0

a is the coefficient of the quadratic term; b is the coefficient of the linear term; c is the value of the constant term.

You can rearrange the equation by collecting all the terms on one side which will result in a 0 on the other side of the equation. When you do this, make sure the quadratic coefficient is positive, and simplify the equation so that the powers of the terms are in descending order.

3b2+14b+40=2b2b2+14b+40=0

The quadratic equation is now in "standard form" and ready to be factorised. Solve for b by making each factor equal to 0, and isolate the variable.

b+10=0b=10b+4=0b=4

The solutions for this equation are b=10 and b=4.


Submit your answer as: and

Quadratic equations: Standard form

Solve for a if:

a2+17a+73=1

You may type your answers in any order.

Answer: a = or
a =
numeric
numeric
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Standard form for a quadratic equation looks like this:

0=ax2+bx+corax2+bx+c=0

You just need to read the values of a, b, and c from the equation.


STEP: Rearrange into standard form and factorise
[−4 points ⇒ 0 / 4 points left]

The equation might look confusing because it is not in standard form. In "standard form," the values of a, b, and c are as follows:

0=ax2+bx+corax2+bx+c=0

a is the coefficient of the quadratic term; b is the coefficient of the linear term; c is the value of the constant term.

You can rearrange the equation by collecting all the terms on one side which will result in a 0 on the other side of the equation. When you do this, make sure the quadratic coefficient is positive, and simplify the equation so that the powers of the terms are in descending order.

a2+17a+73=1a2+17a+72=0

The quadratic equation is now in "standard form" and ready to be factorised. Solve for a by making each factor equal to 0, and isolate the variable.

a+9=0a=9a+8=0a=8

The solutions for this equation are a=9 and a=8.


Submit your answer as: and

Solving quadratic equations

Solve for m. You can type your answers in any order.

4m224m+28=0
Answer: m= or m=
numeric
numeric
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

You need to solve this equation by factorisation, but there is a step you can do before you factorise which will help make the problem easier!


STEP: <no title>
[−1 point ⇒ 3 / 4 points left]

All of the terms in this equation have a factor of 4. Divide the equation by 4 to remove this common factor and simplify the equation. Factorisation is easier with smaller numbers!

m2+6m7=0

STEP: <no title>
[−1 point ⇒ 2 / 4 points left]

Factorise the left hand side:

(m1)(m+7)=0

STEP: <no title>
[−2 points ⇒ 0 / 4 points left]

From the two binomials, you get two possibilities: either m1=0 or m+7=0. From these two options, you get the answers: m=7 or m=1.


Submit your answer as: and

Solving quadratic equations

Solve for a. You can type your answers in any order.

0=2a2+12a16
Answer: a= or a=
numeric
numeric
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

You need to solve this equation by factorisation, but there is a step you can do before you factorise which will help make the problem easier!


STEP: <no title>
[−1 point ⇒ 3 / 4 points left]

All of the terms in this equation have a factor of 2. Divide the equation by 2 to remove this common factor and simplify the equation. Factorisation is easier with smaller numbers!

0=a26a+8

STEP: <no title>
[−1 point ⇒ 2 / 4 points left]

Factorise the right hand side:

0=(a4)(a2)

STEP: <no title>
[−2 points ⇒ 0 / 4 points left]

From the two binomials, you get two possibilities: either a4=0 or a2=0. From these two options, you get the answers: a=2 or a=4.


Submit your answer as: and

Solving quadratic equations

Solve for z. You can type your answers in any order.

0=2z28z+6
Answer: z= or z=
numeric
numeric
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

You need to solve this equation by factorisation, but there is a step you can do before you factorise which will help make the problem easier!


STEP: <no title>
[−1 point ⇒ 3 / 4 points left]

All of the terms in this equation have a factor of 2. Divide the equation by 2 to remove this common factor and simplify the equation. Factorisation is easier with smaller numbers!

0=z24z+3

STEP: <no title>
[−1 point ⇒ 2 / 4 points left]

Factorise the right hand side:

0=(z3)(z1)

STEP: <no title>
[−2 points ⇒ 0 / 4 points left]

From the two binomials, you get two possibilities: either z3=0 or z1=0. From these two options, you get the answers: z=1 or z=3.


Submit your answer as: and

Equations with trinomials

Solve for a if:

a2+11a+18=0

You can type your answers in any order.

Answer: a = or a =
numeric
numeric
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

STEP: <no title>
[−2 points ⇒ 0 / 2 points left]

The first step in solving this problem is to factorise the trinomial. Once you have determined the two factors, apply the zero product rule by setting each factor equal to zero. You can then solve the two factors by isolating the variable, a.

a2+11a+18=0
(a+9)(a+2)=0
a+9=0a+2=0a=9a=2

a=9 or a=2.


Submit your answer as: and

Equations with trinomials

Solve for b if:

b211b+18=0

You can type your answers in any order.

Answer: b = or b =
numeric
numeric
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

STEP: <no title>
[−2 points ⇒ 0 / 2 points left]

The first step in solving this problem is to factorise the trinomial. Once you have determined the two factors, apply the zero product rule by setting each factor equal to zero. You can then solve the two factors by isolating the variable, b.

b211b+18=0
(b9)(b2)=0
b9=0b2=0b=9b=2

b=2 or b=9.


Submit your answer as: and

Equations with trinomials

Solve for x if:

x210x+24=0

You can type your answers in any order.

Answer: x = or x =
numeric
numeric
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

STEP: <no title>
[−2 points ⇒ 0 / 2 points left]

The first step in solving this problem is to factorise the trinomial. Once you have determined the two factors, apply the zero product rule by setting each factor equal to zero. You can then solve the two factors by isolating the variable, x.

x210x+24=0
(x4)(x6)=0
x4=0x6=0x=4x=6

x=4 or x=6.


Submit your answer as: and

Equations with trinomials

Solve for b if:

3b2+10b+8=0

You may type your answers in any order.

Answer: b = or b =
numeric
numeric
STEP: <no title>
[−2 points ⇒ 0 / 2 points left]

The first step in solving this problem is to factorise the trinomial.

3b2+10b+8=0(3b+4)(b+2)=0

Once we have the two binomial factors, we use the zero product rule by setting each factor equal to zero. This leads to two separate equations. Then we can solve each of the equations by isolating the variable, b.

3b+4=0b+2=0b=43b=2

The correct answers are: b=2 or b=43.


Submit your answer as: and

Equations with trinomials

Solve for z if:

10z2+19z+6=0

You may type your answers in any order.

Answer: z = or z =
numeric
numeric
STEP: <no title>
[−2 points ⇒ 0 / 2 points left]

The first step in solving this problem is to factorise the trinomial.

10z2+19z+6=0(5z+2)(2z+3)=0

Once we have the two binomial factors, we use the zero product rule by setting each factor equal to zero. This leads to two separate equations. Then we can solve each of the equations by isolating the variable, z.

5z+2=02z+3=0z=25z=32

The correct answers are: z=32 or z=25.


Submit your answer as: and

Equations with trinomials

Solve for x if:

3x2+13x+10=0

You may type your answers in any order.

Answer: x = or x =
numeric
numeric
STEP: <no title>
[−2 points ⇒ 0 / 2 points left]

The first step in solving this problem is to factorise the trinomial.

3x2+13x+10=0(3x+10)(x+1)=0

Once we have the two binomial factors, we use the zero product rule by setting each factor equal to zero. This leads to two separate equations. Then we can solve each of the equations by isolating the variable, x.

3x+10=0x+1=0x=103x=1

The correct answers are: x=103 or x=1.


Submit your answer as: and

Standard form for quadratic equations

Write this equation in standard form.

(3f+4)(5f2)=13
INSTRUCTION: Write only the left side of the equation in the answer box.
Answer:

=0

polynomial
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Standard form for a quadratic equation is ax2+bx+c=0. You need to multiply the brackets and then simplify so that you can arrange the equation so that it matches standard form.


STEP: Expand the brackets and collect like terms
[−2 points ⇒ 1 / 3 points left]

We must expand the brackets. Also, remember to bring the constant over from the other side of the equation because standard form for a quadratic equation requires a zero on one side of the equation:

(3f+4)(5f2)=1315f26f+20f8+13=0

STEP: Simplify the equation
[−1 point ⇒ 0 / 3 points left]

Now simplify the expression on the left of the equation by collecting like terms.

15f26f+20f8+13=015f2+14f+5=0

The equation now matches the arrangement for standard form:af2+bf+c=0; so we are done!

The final answer is 15f2+14f+5=0


Submit your answer as:

Standard form for quadratic equations

Multiply the brackets and then write the equation in standard form.

(f+4)(2f+5)=5
INSTRUCTION: Write only the left side of the equation in the answer box.
Answer:

=0

polynomial
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Standard form for a quadratic equation is ax2+bx+c=0. You need to multiply the brackets and then simplify so that you can arrange the equation so that it matches standard form.


STEP: Expand the brackets and collect like terms
[−2 points ⇒ 1 / 3 points left]

We must expand the brackets. Also, remember to bring the constant over from the other side of the equation because standard form for a quadratic equation requires a zero on one side of the equation:

(f+4)(2f+5)=52f2+5f+8f+205=0

STEP: Simplify the equation
[−1 point ⇒ 0 / 3 points left]

Now simplify the expression on the left of the equation by collecting like terms.

2f2+5f+8f+205=02f2+13f+15=0

The equation now matches the arrangement for standard form:af2+bf+c=0; so we are done!

The final answer is 2f2+13f+15=0


Submit your answer as:

Standard form for quadratic equations

Write this equation in standard form.

(k+4)(k+4)=13
INSTRUCTION: Write only the left side of the equation in the answer box.
Answer:

=0

polynomial
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Standard form for a quadratic equation is ax2+bx+c=0. You need to multiply the brackets and then simplify so that you can arrange the equation so that it matches standard form.


STEP: Expand the brackets and collect like terms
[−2 points ⇒ 1 / 3 points left]

We must expand the brackets. Also, remember to bring the constant over from the other side of the equation because standard form for a quadratic equation requires a zero on one side of the equation:

(k+4)(k+4)=13k2+4k+4k+1613=0

STEP: Simplify the equation
[−1 point ⇒ 0 / 3 points left]

Now simplify the expression on the left of the equation by collecting like terms.

k2+4k+4k+1613=0k2+8k+3=0

The equation now matches the arrangement for standard form:ak2+bk+c=0; so we are done!

The final answer is k2+8k+3=0


Submit your answer as:

Equations with a difference of two squares

Solve for z if:

z2100=0

You can type your answers in any order.

Answer: z = or z =
numeric
numeric
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

STEP: <no title>
[−2 points ⇒ 0 / 2 points left]

The first step in solving this problem is to factorise the binomial. This binomial is the the difference of two squares. When you have the two factors you set them equal to zero, applying the zero product rule. You then solve the two factors by isolating the variable, z.

z2100=0
(z+10)(z10)=0
z+10=0z10=0z=10z=10

z=10 or z=10


Submit your answer as: and

Equations with a difference of two squares

Solve for x if:

x225=0

You can type your answers in any order.

Answer: x = or x =
numeric
numeric
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

STEP: <no title>
[−2 points ⇒ 0 / 2 points left]

The first step in solving this problem is to factorise the binomial. This binomial is the the difference of two squares. When you have the two factors you set them equal to zero, applying the zero product rule. You then solve the two factors by isolating the variable, x.

x225=0
(x+5)(x5)=0
x+5=0x5=0x=5x=5

x=5 or x=5


Submit your answer as: and

Equations with a difference of two squares

Solve for x if:

x236=0

You can type your answers in any order.

Answer: x = or x =
numeric
numeric
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

STEP: <no title>
[−2 points ⇒ 0 / 2 points left]

The first step in solving this problem is to factorise the binomial. This binomial is the the difference of two squares. When you have the two factors you set them equal to zero, applying the zero product rule. You then solve the two factors by isolating the variable, x.

x236=0
(x+6)(x6)=0
x+6=0x6=0x=6x=6

x=6 or x=6


Submit your answer as: and

Equations with trinomials

Solve for x if:

x2+4x+3=0

You can type your answers in any order.

Answer: x = or x =
numeric
numeric
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

STEP: <no title>
[−2 points ⇒ 0 / 2 points left]

The first step in solving this problem is to factorise the trinomial. When you have the two factors you set them equal to zero, applying the zero product rule. You then solve the two factors by isolating the variable, x.

x2+4x+3=0
(x+1)(x+3)=0
x+1=0x+3=0x=1x=3

x=3 or x=1.


Submit your answer as: and

Equations with trinomials

Solve for y if:

y2+5y+4=0

You can type your answers in any order.

Answer: y = or y =
numeric
numeric
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

STEP: <no title>
[−2 points ⇒ 0 / 2 points left]

The first step in solving this problem is to factorise the trinomial. When you have the two factors you set them equal to zero, applying the zero product rule. You then solve the two factors by isolating the variable, y.

y2+5y+4=0
(y+4)(y+1)=0
y+4=0y+1=0y=4y=1

y=4 or y=1.


Submit your answer as: and

Equations with trinomials

Solve for a if:

a2+14a+45=0

You can type your answers in any order.

Answer: a = or a =
numeric
numeric
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

STEP: <no title>
[−2 points ⇒ 0 / 2 points left]

The first step in solving this problem is to factorise the trinomial. When you have the two factors you set them equal to zero, applying the zero product rule. You then solve the two factors by isolating the variable, a.

a2+14a+45=0
(a+9)(a+5)=0
a+9=0a+5=0a=9a=5

a=9 or a=5.


Submit your answer as: and

Equations with trinomials

Solve for a if:

5a2+36a81=0

You can type your answers in any order.

Answer: a = or a =
numeric
numeric
STEP: <no title>
[−2 points ⇒ 0 / 2 points left]

The first step in solving this problem is to factorise the trinomial. Once we have determined the two factors, we use the zero product rule by setting each factor equal to zero. We can then solve the two factors by isolating the variable, a.

5a2+36a81=0(a+9)(5a9)=0
a+9=05a9=0a=9a=95

The solutions to the equation are a=9 and a=95.


Submit your answer as: and

Equations with trinomials

Solve for b if:

3b2+20b100=0

You can type your answers in any order.

Answer: b = or b =
numeric
numeric
STEP: <no title>
[−2 points ⇒ 0 / 2 points left]

The first step in solving this problem is to factorise the trinomial. Once we have determined the two factors, we use the zero product rule by setting each factor equal to zero. We can then solve the two factors by isolating the variable, b.

3b2+20b100=0(3b10)(b+10)=0
3b10=0b+10=0b=103b=10

The solutions to the equation are b=10 and b=103.


Submit your answer as: and

Equations with trinomials

Solve for a if:

9a29a70=0

You can type your answers in any order.

Answer: a = or a =
numeric
numeric
STEP: <no title>
[−2 points ⇒ 0 / 2 points left]

The first step in solving this problem is to factorise the trinomial. Once we have determined the two factors, we use the zero product rule by setting each factor equal to zero. We can then solve the two factors by isolating the variable, a.

9a29a70=0(3a+7)(3a10)=0
3a+7=03a10=0a=73a=103

The solutions to the equation are a=73 and a=103.


Submit your answer as: and

Solve quadratic equations by inspection

Solve for a:

(a+1)2=64
Answer: a= or a=
fraction
fraction
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Which two numbers will give 64 when squared? Remember, when you multiply a negative number by itself you get a positive number.


STEP: Determine two numbers that give 64 when squared
[−2 points ⇒ 1 / 3 points left]

We already know how to solve linear equations using inverse operations. But, this approach does not always work for non-linear equations. This equation has a a2 term, so it is not linear (it is quadratic). We use inspection to solve non-linear equations.

We will start by solving this equation for the whole bracket:

(a+1)2=64

To answer this question we ask "What number (or numbers), when you square it, gives you 64?".

Since 8×8=64, we know that 8 is a value for a+1.

However, this is not the only number that gives 64 when multiplied by itself. Remember, when you square a negative number you get a positive number:

(8)2=(8)(8)=64

Therefore, a+1 must be equal to 8 or 8.


STEP: Determine a
[−1 point ⇒ 0 / 3 points left]

Mathematically we write this as:

(a+1)2=64a+1=8 or a+1=8

Now we have created two linear equations that can each be solved separately for a, using inverse operations.

a+1=8a+11=81a=7

or

a+1=8a+11=81a=9

We can check our answers by substituting in the values for a:

  • If a=7: (a+1)2 = (7+1)2 = (8)2 = 64
  • If a=9: (a+1)2 = (9+1)2 = (8)2 = 64

Since these a-values make the equation true, we know that they are both valid answers.


Submit your answer as: and

Solve quadratic equations by inspection

Solve for p:

(p5)2=1
Answer: p= or p=
fraction
fraction
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Which two numbers will give 1 when squared? Remember, when you multiply a negative number by itself you get a positive number.


STEP: Determine two numbers that give 1 when squared
[−2 points ⇒ 1 / 3 points left]

We already know how to solve linear equations using inverse operations. But, this approach does not always work for non-linear equations. This equation has a p2 term, so it is not linear (it is quadratic). We use inspection to solve non-linear equations.

We will start by solving this equation for the whole bracket:

(p5)2=1

To answer this question we ask "What number (or numbers), when you square it, gives you 1?".

Since 1×1=1, we know that 1 is a value for p5.

However, this is not the only number that gives 1 when multiplied by itself. Remember, when you square a negative number you get a positive number:

(1)2=(1)(1)=1

Therefore, p5 must be equal to 1 or 1.


STEP: Determine p
[−1 point ⇒ 0 / 3 points left]

Mathematically we write this as:

(p5)2=1p5=1 or p5=1

Now we have created two linear equations that can each be solved separately for p, using inverse operations.

p5=1p5+5=1+5p=6

or

p5=1p5+5=1+5p=4

We can check our answers by substituting in the values for p:

  • If p=6: (p5)2 = (65)2 = (1)2 = 1
  • If p=4: (p5)2 = (45)2 = (1)2 = 1

Since these p-values make the equation true, we know that they are both valid answers.


Submit your answer as: and

Solve quadratic equations by inspection

Solve for p:

(p+3)2=16
Answer: p= or p=
fraction
fraction
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Which two numbers will give 16 when squared? Remember, when you multiply a negative number by itself you get a positive number.


STEP: Determine two numbers that give 16 when squared
[−2 points ⇒ 1 / 3 points left]

We already know how to solve linear equations using inverse operations. But, this approach does not always work for non-linear equations. This equation has a p2 term, so it is not linear (it is quadratic). We use inspection to solve non-linear equations.

We will start by solving this equation for the whole bracket:

(p+3)2=16

To answer this question we ask "What number (or numbers), when you square it, gives you 16?".

Since 4×4=16, we know that 4 is a value for p+3.

However, this is not the only number that gives 16 when multiplied by itself. Remember, when you square a negative number you get a positive number:

(4)2=(4)(4)=16

Therefore, p+3 must be equal to 4 or 4.


STEP: Determine p
[−1 point ⇒ 0 / 3 points left]

Mathematically we write this as:

(p+3)2=16p+3=4 or p+3=4

Now we have created two linear equations that can each be solved separately for p, using inverse operations.

p+3=4p+33=43p=1

or

p+3=4p+33=43p=7

We can check our answers by substituting in the values for p:

  • If p=1: (p+3)2 = (1+3)2 = (4)2 = 16
  • If p=7: (p+3)2 = (7+3)2 = (4)2 = 16

Since these p-values make the equation true, we know that they are both valid answers.


Submit your answer as: and

3. Creating quadratic equations

Finding a quadratic equation from its roots

  1. Determine an equation which has roots at x=14 and x=9.

    INSTRUCTION: Write the equation in standard form: ax2+bx+c=0 where a,b,cZ.
    Answer: Write the entire equation here:
    one-of
    type(equation)
    HINT: <no title>
    [−0 points ⇒ 4 / 4 points left]

    To find the equation, start with the basic form here: (xroot1)(xroot2). When fractions are involved in this type of question, remember that you will want one of the brackets to take the form (axb), where a is the denominator of the fraction and b is the numerator.


    STEP: Use the roots to find the factors
    [−2 points ⇒ 2 / 4 points left]

    To solve this question, it is like we are solving a quadratic equation backwards: we have the roots of an equation, and now we want to find that equation. Remember that roots are values that make the quadratic expression equal to zero.

    Let's start with the two roots, but rearrange them as follows:

    x=14x(14)=0x14=0

    and

    x=9x(9)=0x+9=0

    STEP: Rearrange the fraction
    [−1 point ⇒ 1 / 4 points left]

    Now we can do something with the fraction to simplify our lives: multiply both sides of the equation by the denominator of the fraction.

    4(x14)=0(4)4x1=0

    and

    x+9=0

    STEP: Multiply the factors to find the equation
    [−1 point ⇒ 0 / 4 points left]

    Now that the fractions are gone, put these two pieces together and multiply out the brackets to get the equation.

    (4x1)(x+9)=04x2+35x9=0
    TIP:

    We can check the answer to make sure that it is correct. Remember that the question tells us that the equation we want has a root equal to 14. Therefore, if x=14, the equation must be equal to zero. If x=14:

    4x2+35x94(14)2+35(14)9=0

    You can try the same with the second root, and you will find the answer comes out to be zero again!

    The correct answer is 4x2+35x9=0.


    Submit your answer as:
  2. Which of the following equations has the same roots as the equation from the first question?

    Answer:
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    You do not need to solve the equations to find their roots. Nor do you need to check the roots in each of the equations. You just need to compare the equations shown to the original equation, and remember that zero multiplied by anything is always zero.


    STEP: Identify the equation which is a multiple of the first equation
    [−1 point ⇒ 0 / 1 points left]

    The correct choice is Equation B. The reason is because each term is a multiple of the matching term in the original equation. If we divide both sides of the equation by 3, then we get the original equation!

    12x2+105x27=012x2+105x273=034x2+35x9=0

    The correct choice is Equation B.


    Submit your answer as:

Completing an equation by using a root.

Determine the value of j in the equation y=jx27x30 given that one of the roots is x=6. Then determine the other root.

Answer: j= and the second root is x=
numeric
numeric
HINT: <no title>
[−1 point ⇒ 4 / 5 points left]
The keys to getting started are these two things: (a) the root value will make the equation equal to zero (meaning that y=0) and (b) when you know values in an equation, you should substitute those values into the equation to see if you can then get the value of j.
STEP: <no title>
[−1 point ⇒ 3 / 5 points left]

You know that x=6 is a root of the equation; that means that when x=6,y=0. Substitute these numbers into the equation, and then tidy up as much as you can:

y=jx27x300=j(6)27(6)300=36j72

STEP: <no title>
[−1 point ⇒ 2 / 5 points left]

Now just solve for j.

0=36j7272=36j2=j

The value of j is 2.


STEP: <no title>
[−1 point ⇒ 1 / 5 points left]

Now you can move on to the second part of the question: finding the other root of the equation. First, you can now write the full equation, becuase you know that j=2:

jx27x302x27x30

STEP: <no title>
[−1 point ⇒ 0 / 5 points left]

We want to find a root, which means (as always!) y=0. To solve for the root, you need to factorise - and it is very helpful to remember that you already know one of the roots (x=6) !

0=2x27x30=(?x+?)(x6)use the root you already know!

By comparing these two lines you can see that the other binomial must contain a 2 and a 5...

0=(2x+5)(x6)

STEP: <no title>
[−1 point ⇒ 0 / 5 points left]

Now just solve for x using the first binomial to get the other root.

2x+5=02x=5x=52

The second root is x=52.


Submit your answer as: and

Finding a quadratic function

The graph of g(x)=ax2+q is sketched below. Points R(3;0) and S(5;8) lie on the graph of g. Points R and T are x-intercepts of g.

  1. What are the coordinates of T?

    Answer: The coordinates of T are .
    coordinate
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Use symmetry to answer this question.


    STEP: Use the symmetry of the quadratic
    [−1 point ⇒ 0 / 1 points left]

    We have the x-intercept R(3;0) and we need the coordinates of the other x-intercept T.

    A quadratic is symmetrical about a vertical line passing through its turning point. This means that the coordinates we have and the coordinates we are looking for are reflections of each other about this line of symmetry (mirror line). In this case, the mirror line happens to be the y-axis.

    A reflection about the y-axis follows the transformation rule

    (x;y)(x;y)

    The sign of the x-coordinate changes and the y-coordinate (zero in this case) stays the same.

    Therefore the coordinates of T are (3;0).


    Submit your answer as:
  2. Determine the equation of g.

    INSTRUCTION: Type only the right side of the equation. Do not type 'g(x)' as part of your answer.
    Answer:

    The equation for g is g(x)= .

    expression
    HINT: <no title>
    [−0 points ⇒ 4 / 4 points left]

    Use the coordinates of the points given to set up simultaneous equations to determine the values of the parameters a and q.

    Then write the equation for g in the form

    g(x)=ax2+q

    As your answer you will type ax2+q with the particular values of a and q.


    STEP: Substitute known points in the equation
    [−2 points ⇒ 2 / 4 points left]

    In order to find the equation of g we need to find values for the two unknowns a and q.

    We have three points on the graph of g for which we have coordinates. These points are S(5;8), and the two x-intercepts R(3;0) and T(3;0).

    If we substitute the coordinates for S and R into the equation for g, we can set up two simultaneous equations to solve for the two unknowns.

    NOTE:

    Instead of Point R, we could also use the other x-intercept, T, now that we have found it in Question 1. Sometimes it is safest to use a given value instead of a value that we have calculated, just in case we have made a mistake.

    We cannot use Points R and T together, as we need two independent points to give us our simultaneous equations. Points R and T are tied together through symmetry. So we must use S with only one of the other x-intercepts.

    Substituting S(5;8) into g gives us:

    8=(5)2a+q8=25a+q(1)

    Substituting R(3;0) into g gives us:

    0=(3)2a+q0=9a+q(2)

    STEP: Solve the simultaneous equations for a and q
    [−2 points ⇒ 0 / 4 points left]

    We can eliminate q from the simultaneous equations by subtracting one equation from the other. This allows us to solve for a first.

    Subtracting Equation 1 from Equation 2 gives us:

    80=(25a+q)(9a+q)8=25a9a8=16a12=a

    We can substitute the value for a that we have just found into either of the equations. Substituting a into Equation 2 gives us:

    0=912+q92=q

    Therefore the equation for g is

    g(x)=x2292

    As your answer you would have entered only the expression on the right hand side:

    x2292

    Submit your answer as:

Finding a quadratic equation from its roots

Given that a quadratic equation has the roots -10 and 10, find the equation.

INSTRUCTION: Write the equation in standard form: x2+bx+c=0. (The quadratic coefficient should be 1.)
Answer: Write the entire equation here: .
one-of
type(equation.strict)
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

To find the equation, you should start with the basic form here: (xroot1)(xroot2).


STEP: Find the factors for the equation
[−2 points ⇒ 1 / 3 points left]

To solve this equation, it is like we are solving a quadratic equation backwards: we have the roots of an equation, and now we want to find that equation. Remember that roots are values that make the quadratic expression equal to zero.

Let's take the two super small equations x=10 and x=10 and then subtract the roots over to the left side to get zeroes (be careful with the signs... in this case there are some negatives which become addition):

x=10x(10)=0x+10=0
x=10x(10)=0x10=0

STEP: Find the equation
[−1 point ⇒ 0 / 3 points left]

We have two expressions above ((x+10) and (x10)) which are both equal to zero; if we multiply them, the answer must be zero. We can multiply the two expressions to get the equation.

0=(x+10)(x10)0=x2100
TIP:

We can check the answer to make sure that it is correct. Remember that the question tells us that the equation we want has a root equal to 10. Therefore, if x=10, the equation must be equal to zero. If x=10:

x2100(10)2100=0

You can try the same with the second root, and you will find the answer comes out to be zero again!

The correct answer is x2100=0.


Submit your answer as:

Exercises

Finding a quadratic equation from its roots

  1. Determine an equation which has roots at x=12 and x=9.

    INSTRUCTION: Write the equation in standard form: ax2+bx+c=0 where a,b,cZ.
    Answer: Write the entire equation here:
    one-of
    type(equation)
    HINT: <no title>
    [−0 points ⇒ 4 / 4 points left]

    To find the equation, start with the basic form here: (xroot1)(xroot2). When fractions are involved in this type of question, remember that you will want one of the brackets to take the form (axb), where a is the denominator of the fraction and b is the numerator.


    STEP: Use the roots to find the factors
    [−2 points ⇒ 2 / 4 points left]

    To solve this question, it is like we are solving a quadratic equation backwards: we have the roots of an equation, and now we want to find that equation. Remember that roots are values that make the quadratic expression equal to zero.

    Let's start with the two roots, but rearrange them as follows:

    x=12x(12)=0x+12=0

    and

    x=9x(9)=0x9=0

    STEP: Rearrange the fraction
    [−1 point ⇒ 1 / 4 points left]

    Now we can do something with the fraction to simplify our lives: multiply both sides of the equation by the denominator of the fraction.

    2(x+12)=0(2)2x+1=0

    and

    x9=0

    STEP: Multiply the factors to find the equation
    [−1 point ⇒ 0 / 4 points left]

    Now that the fractions are gone, put these two pieces together and multiply out the brackets to get the equation.

    (2x+1)(x9)=02x217x9=0
    TIP:

    We can check the answer to make sure that it is correct. Remember that the question tells us that the equation we want has a root equal to 12. Therefore, if x=12, the equation must be equal to zero. If x=12:

    2x217x92(12)217(12)9=0

    You can try the same with the second root, and you will find the answer comes out to be zero again!

    The correct answer is 2x217x9=0.


    Submit your answer as:
  2. Which of the following equations has the same roots as the equation from the first question?

    Answer:
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    You do not need to solve the equations to find their roots. Nor do you need to check the roots in each of the equations. You just need to compare the equations shown to the original equation, and remember that zero multiplied by anything is always zero.


    STEP: Identify the equation which is a multiple of the first equation
    [−1 point ⇒ 0 / 1 points left]

    The correct choice is Equation A. The reason is because each term is a multiple of the matching term in the original equation. If we divide both sides of the equation by 2, then we get the original equation!

    4x2+34x+18=04x2+34x+182=022x217x9=0

    The correct choice is Equation A.


    Submit your answer as:

Finding a quadratic equation from its roots

  1. Given that a quadratic equation has the roots 73 and 9, find the equation.

    INSTRUCTION: Write the equation in standard form: ax2+bx+c=0 where a,b,cZ.
    Answer: Write the entire equation here:
    one-of
    type(equation)
    HINT: <no title>
    [−0 points ⇒ 4 / 4 points left]

    To find the equation, start with the basic form here: (xroot1)(xroot2). When fractions are involved in this type of question, remember that you will want one of the brackets to take the form (axb), where a is the denominator of the fraction and b is the numerator.


    STEP: Use the roots to find the factors
    [−2 points ⇒ 2 / 4 points left]

    To solve this question, it is like we are solving a quadratic equation backwards: we have the roots of an equation, and now we want to find that equation. Remember that roots are values that make the quadratic expression equal to zero.

    Let's start with the two roots, but rearrange them as follows:

    x=73x(73)=0x+73=0

    and

    x=9x(9)=0x9=0

    STEP: Rearrange the fraction
    [−1 point ⇒ 1 / 4 points left]

    Now we can do something with the fraction to simplify our lives: multiply both sides of the equation by the denominator of the fraction.

    3(x+73)=0(3)3x+7=0

    and

    x9=0

    STEP: Multiply the factors to find the equation
    [−1 point ⇒ 0 / 4 points left]

    Now that the fractions are gone, put these two pieces together and multiply out the brackets to get the equation.

    (3x+7)(x9)=03x220x63=0
    TIP:

    We can check the answer to make sure that it is correct. Remember that the question tells us that the equation we want has a root equal to 73. Therefore, if x=73, the equation must be equal to zero. If x=73:

    3x220x633(73)220(73)63=0

    You can try the same with the second root, and you will find the answer comes out to be zero again!

    The correct answer is 3x220x63=0.


    Submit your answer as:
  2. Which of the following equations has the same roots as the equation from the first question?

    Answer:
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    You do not need to solve the equations to find their roots. Nor do you need to check the roots in each of the equations. You just need to compare the equations shown to the original equation, and remember that zero multiplied by anything is always zero.


    STEP: Identify the equation which is a multiple of the first equation
    [−1 point ⇒ 0 / 1 points left]

    The correct choice is Equation D. The reason is because each term is a multiple of the matching term in the original equation. If we divide both sides of the equation by 1, then we get the original equation!

    3x2+20x+63=03x2+20x+631=013x220x63=0

    The correct choice is Equation D.


    Submit your answer as:

Finding a quadratic equation from its roots

  1. Find an equation with roots 72 and 5.

    INSTRUCTION: Write the equation in standard form: ax2+bx+c=0 where a,b,cZ.
    Answer: Write the entire equation here:
    one-of
    type(equation)
    HINT: <no title>
    [−0 points ⇒ 4 / 4 points left]

    To find the equation, start with the basic form here: (xroot1)(xroot2). When fractions are involved in this type of question, remember that you will want one of the brackets to take the form (axb), where a is the denominator of the fraction and b is the numerator.


    STEP: Use the roots to find the factors
    [−2 points ⇒ 2 / 4 points left]

    To solve this question, it is like we are solving a quadratic equation backwards: we have the roots of an equation, and now we want to find that equation. Remember that roots are values that make the quadratic expression equal to zero.

    Let's start with the two roots, but rearrange them as follows:

    x=72x(72)=0x+72=0

    and

    x=5x(5)=0x5=0

    STEP: Rearrange the fraction
    [−1 point ⇒ 1 / 4 points left]

    Now we can do something with the fraction to simplify our lives: multiply both sides of the equation by the denominator of the fraction.

    2(x+72)=0(2)2x+7=0

    and

    x5=0

    STEP: Multiply the factors to find the equation
    [−1 point ⇒ 0 / 4 points left]

    Now that the fractions are gone, put these two pieces together and multiply out the brackets to get the equation.

    (2x+7)(x5)=02x23x35=0
    TIP:

    We can check the answer to make sure that it is correct. Remember that the question tells us that the equation we want has a root equal to 72. Therefore, if x=72, the equation must be equal to zero. If x=72:

    2x23x352(72)23(72)35=0

    You can try the same with the second root, and you will find the answer comes out to be zero again!

    The correct answer is 2x23x35=0.


    Submit your answer as:
  2. Which of the following equations has the same roots as the equation from the first question?

    Answer:
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    You do not need to solve the equations to find their roots. Nor do you need to check the roots in each of the equations. You just need to compare the equations shown to the original equation, and remember that zero multiplied by anything is always zero.


    STEP: Identify the equation which is a multiple of the first equation
    [−1 point ⇒ 0 / 1 points left]

    The correct choice is Equation D. The reason is because each term is a multiple of the matching term in the original equation. If we divide both sides of the equation by 3, then we get the original equation!

    6x29x105=06x29x1053=032x23x35=0

    The correct choice is Equation D.


    Submit your answer as:

Completing an equation by using a root.

Determine the value of m in the equation y=mx2+9x+2 given that one of the roots is x=2. Then determine the other root.

Answer: m= and the second root is x=
numeric
numeric
HINT: <no title>
[−1 point ⇒ 4 / 5 points left]
The keys to getting started are these two things: (a) the root value will make the equation equal to zero (meaning that y=0) and (b) when you know values in an equation, you should substitute those values into the equation to see if you can then get the value of m.
STEP: <no title>
[−1 point ⇒ 3 / 5 points left]

You know that x=2 is a root of the equation; that means that when x=2,y=0. Substitute these numbers into the equation, and then tidy up as much as you can:

y=mx2+9x+20=m(2)2+9(2)+20=4m16

STEP: <no title>
[−1 point ⇒ 2 / 5 points left]

Now just solve for m.

0=4m1616=4m4=m

The value of m is 4.


STEP: <no title>
[−1 point ⇒ 1 / 5 points left]

Now you can move on to the second part of the question: finding the other root of the equation. First, you can now write the full equation, becuase you know that m=4:

mx2+9x+24x2+9x+2

STEP: <no title>
[−1 point ⇒ 0 / 5 points left]

We want to find a root, which means (as always!) y=0. To solve for the root, you need to factorise - and it is very helpful to remember that you already know one of the roots (x=2) !

0=4x2+9x+2=(?x+?)(x+2)use the root you already know!

By comparing these two lines you can see that the other binomial must contain a 4 and a 1...

0=(4x+1)(x+2)

STEP: <no title>
[−1 point ⇒ 0 / 5 points left]

Now just solve for x using the first binomial to get the other root.

4x+1=04x=1x=14

The second root is x=14.


Submit your answer as: and

Completing an equation by using a root.

Determine the value of m in the equation y=2x2+9x+m given that one of the roots is x=3. Then determine the other root.

Answer: m= and the second root is x=
numeric
numeric
HINT: <no title>
[−1 point ⇒ 4 / 5 points left]
The keys to getting started are these two things: (a) the root value will make the equation equal to zero (meaning that y=0) and (b) when you know values in an equation, you should substitute those values into the equation to see if you can then get the value of m.
STEP: <no title>
[−1 point ⇒ 3 / 5 points left]

You know that x=3 is a root of the equation; that means that when x=3,y=0. Substitute these numbers into the equation, and then tidy up as much as you can:

y=2x2+9x+m0=2(3)2+9(3)+m0=m9

STEP: <no title>
[−1 point ⇒ 2 / 5 points left]

Now just solve for m.

0=m99=m

The value of m is 9.


STEP: <no title>
[−1 point ⇒ 1 / 5 points left]

Now you can move on to the second part of the question: finding the other root of the equation. First, you can now write the full equation, becuase you know that m=9:

2x2+9x+m2x2+9x+9

STEP: <no title>
[−1 point ⇒ 0 / 5 points left]

We want to find a root, which means (as always!) y=0. To solve for the root, you need to factorise - and it is very helpful to remember that you already know one of the roots (x=3) !

0=2x2+9x+9=(?x+?)(x+3)use the root you already know!

By comparing these two lines you can see that the other binomial must contain a 2 and a 3...

0=(2x+3)(x+3)

STEP: <no title>
[−1 point ⇒ 0 / 5 points left]

Now just solve for x using the first binomial to get the other root.

2x+3=02x=3x=32

The second root is x=32.


Submit your answer as: and

Completing an equation by using a root.

Determine the value of m in the equation y=2x2+x+m given that one of the roots is x=2. Then determine the other root.

Answer: m= and the second root is x=
numeric
numeric
HINT: <no title>
[−1 point ⇒ 4 / 5 points left]
The keys to getting started are these two things: (a) the root value will make the equation equal to zero (meaning that y=0) and (b) when you know values in an equation, you should substitute those values into the equation to see if you can then get the value of m.
STEP: <no title>
[−1 point ⇒ 3 / 5 points left]

You know that x=2 is a root of the equation; that means that when x=2,y=0. Substitute these numbers into the equation, and then tidy up as much as you can:

y=2x2+x+m0=2(2)2+(2)+m0=m+10

STEP: <no title>
[−1 point ⇒ 2 / 5 points left]

Now just solve for m.

0=m+1010=m

The value of m is 10.


STEP: <no title>
[−1 point ⇒ 1 / 5 points left]

Now you can move on to the second part of the question: finding the other root of the equation. First, you can now write the full equation, becuase you know that m=10:

2x2+x+m2x2+x10

STEP: <no title>
[−1 point ⇒ 0 / 5 points left]

We want to find a root, which means (as always!) y=0. To solve for the root, you need to factorise - and it is very helpful to remember that you already know one of the roots (x=2) !

0=2x2+x10=(?x+?)(x2)use the root you already know!

By comparing these two lines you can see that the other binomial must contain a 2 and a 5...

0=(2x+5)(x2)

STEP: <no title>
[−1 point ⇒ 0 / 5 points left]

Now just solve for x using the first binomial to get the other root.

2x+5=02x=5x=52

The second root is x=52.


Submit your answer as: and

Finding a quadratic function

The graph of h(x)=ax2+q is sketched below. Points P(3;0) and Q(1;6) lie on the graph of h. Points P and R are x-intercepts of h.

  1. What are the coordinates of R?

    Answer: The coordinates of R are .
    coordinate
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Use symmetry to answer this question.


    STEP: Use the symmetry of the parabola
    [−1 point ⇒ 0 / 1 points left]

    We have the x-intercept P(3;0) and we need the coordinates of the other x-intercept R.

    A parabola is symmetrical about a vertical line passing through its turning point. This means that the coordinates we have and the coordinates we are looking for are reflections of each other about this line of symmetry (mirror line). In this case, the mirror line happens to be the y-axis.

    A reflection about the y-axis follows the transformation rule

    (x;y)(x;y)

    The sign of the x-coordinate changes and the y-coordinate (zero in this case) stays the same.

    Therefore the coordinates of R are (3;0).


    Submit your answer as:
  2. Determine the equation of h.

    INSTRUCTION: Type only the right side of the equation. Do not type 'h(x)' as part of your answer.
    Answer:

    The equation for h is h(x)= .

    expression
    HINT: <no title>
    [−0 points ⇒ 4 / 4 points left]

    Use the coordinates of the points given to set up simultaneous equations to determine the values of the parameters a and q.

    Then write the equation for h in the form

    h(x)=ax2+q

    As your answer you will type ax2+q with the particular values of a and q.


    STEP: Substitute known points in the equation
    [−2 points ⇒ 2 / 4 points left]

    In order to find the equation of h we need to find values for the two unknowns a and q.

    We have three points on the graph of h for which we have coordinates. These points are Q(1;6), and the two x-intercepts P(3;0) and R(3;0).

    If we substitute the coordinates for Q and P into the equation for h, we can set up two simultaneous equations to solve for the two unknowns.

    NOTE:

    Instead of Point P, we could also use the other x-intercept, R, now that we have found it in Question 1. Sometimes it is safest to use a given value instead of a value that we have calculated, just in case we have made a mistake.

    We cannot use Points P and R together, as we need two independent points to give us our simultaneous equations. Points P and R are tied together through symmetry. So we must use Q with only one of the other x-intercepts.

    Substituting Q(1;6) into h gives us:

    6=(1)2a+q6=a+q(1)

    Substituting P(3;0) into h gives us:

    0=(3)2a+q0=9a+q(2)

    STEP: Solve the simultaneous equations for a and q
    [−2 points ⇒ 0 / 4 points left]

    We can eliminate q from the simultaneous equations by subtracting one equation from the other. This allows us to solve for a first.

    Subtracting Equation 1 from Equation 2 gives us:

    60=(a+q)(9a+q)6=a9a6=8a34=a

    We can substitute the value for a that we have just found into either of the equations. Substituting a into Equation 2 gives us:

    0=934+q274=q

    Therefore the equation for h is

    h(x)=3x24+274

    As your answer you would have entered only the expression on the right hand side:

    3x24+274

    Submit your answer as:

Finding a quadratic function

The graph of f(x)=ax2+q is sketched below. Points P(1;0) and Q(2;3) lie on the graph of f. Points P and R are x-intercepts of f.

  1. What are the coordinates of R?

    Answer: The coordinates of R are .
    coordinate
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Use symmetry to answer this question.


    STEP: Use the symmetry of the quadratic
    [−1 point ⇒ 0 / 1 points left]

    We have the x-intercept P(1;0) and we need the coordinates of the other x-intercept R.

    A quadratic is symmetrical about a vertical line passing through its turning point. This means that the coordinates we have and the coordinates we are looking for are reflections of each other about this line of symmetry (mirror line). In this case, the mirror line happens to be the y-axis.

    A reflection about the y-axis follows the transformation rule

    (x;y)(x;y)

    The sign of the x-coordinate changes and the y-coordinate (zero in this case) stays the same.

    Therefore the coordinates of R are (1;0).


    Submit your answer as:
  2. Determine the equation of f.

    INSTRUCTION: Type only the right side of the equation. Do not type 'f(x)' as part of your answer.
    Answer:

    The equation for f is f(x)= .

    expression
    HINT: <no title>
    [−0 points ⇒ 4 / 4 points left]

    Use the coordinates of the points given to set up simultaneous equations to determine the values of the parameters a and q.

    Then write the equation for f in the form

    f(x)=ax2+q

    As your answer you will type ax2+q with the particular values of a and q.


    STEP: Substitute known points in the equation
    [−2 points ⇒ 2 / 4 points left]

    In order to find the equation of f we need to find values for the two unknowns a and q.

    We have three points on the graph of f for which we have coordinates. These points are Q(2;3), and the two x-intercepts P(1;0) and R(1;0).

    If we substitute the coordinates for Q and P into the equation for f, we can set up two simultaneous equations to solve for the two unknowns.

    NOTE:

    Instead of Point P, we could also use the other x-intercept, R, now that we have found it in Question 1. Sometimes it is safest to use a given value instead of a value that we have calculated, just in case we have made a mistake.

    We cannot use Points P and R together, as we need two independent points to give us our simultaneous equations. Points P and R are tied together through symmetry. So we must use Q with only one of the other x-intercepts.

    Substituting Q(2;3) into f gives us:

    3=(2)2a+q3=4a+q(1)

    Substituting P(1;0) into f gives us:

    0=(1)2a+q0=a+q(2)

    STEP: Solve the simultaneous equations for a and q
    [−2 points ⇒ 0 / 4 points left]

    We can eliminate q from the simultaneous equations by subtracting one equation from the other. This allows us to solve for a first.

    Subtracting Equation 1 from Equation 2 gives us:

    30=(4a+q)(a+q)3=4aa3=3a1=a

    We can substitute the value for a that we have just found into either of the equations. Substituting a into Equation 2 gives us:

    0=11+q1=q

    Therefore the equation for f is

    f(x)=x2+1

    As your answer you would have entered only the expression on the right hand side:

    x2+1

    Submit your answer as:

Finding a quadratic function

The graph of f(x)=ax2+q is sketched below. Points R(1;0) and S(2;6) lie on the graph of f. Points R and T are x-intercepts of f.

  1. What are the coordinates of T?

    Answer: The coordinates of T are .
    coordinate
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Use symmetry to answer this question.


    STEP: Use the symmetry of the parabola
    [−1 point ⇒ 0 / 1 points left]

    We have the x-intercept R(1;0) and we need the coordinates of the other x-intercept T.

    A parabola is symmetrical about a vertical line passing through its turning point. This means that the coordinates we have and the coordinates we are looking for are reflections of each other about this line of symmetry (mirror line). In this case, the mirror line happens to be the y-axis.

    A reflection about the y-axis follows the transformation rule

    (x;y)(x;y)

    The sign of the x-coordinate changes and the y-coordinate (zero in this case) stays the same.

    Therefore the coordinates of T are (1;0).


    Submit your answer as:
  2. Determine the equation of f.

    INSTRUCTION: Type only the right side of the equation. Do not type 'f(x)' as part of your answer.
    Answer:

    The equation for f is f(x)= .

    expression
    HINT: <no title>
    [−0 points ⇒ 4 / 4 points left]

    Use the coordinates of the points given to set up simultaneous equations to determine the values of the parameters a and q.

    Then write the equation for f in the form

    f(x)=ax2+q

    As your answer you will type ax2+q with the particular values of a and q.


    STEP: Substitute known points in the equation
    [−2 points ⇒ 2 / 4 points left]

    In order to find the equation of f we need to find values for the two unknowns a and q.

    We have three points on the graph of f for which we have coordinates. These points are S(2;6), and the two x-intercepts R(1;0) and T(1;0).

    If we substitute the coordinates for S and R into the equation for f, we can set up two simultaneous equations to solve for the two unknowns.

    NOTE:

    Instead of Point R, we could also use the other x-intercept, T, now that we have found it in Question 1. Sometimes it is safest to use a given value instead of a value that we have calculated, just in case we have made a mistake.

    We cannot use Points R and T together, as we need two independent points to give us our simultaneous equations. Points R and T are tied together through symmetry. So we must use S with only one of the other x-intercepts.

    Substituting S(2;6) into f gives us:

    6=(2)2a+q6=4a+q(1)

    Substituting R(1;0) into f gives us:

    0=(1)2a+q0=a+q(2)

    STEP: Solve the simultaneous equations for a and q
    [−2 points ⇒ 0 / 4 points left]

    We can eliminate q from the simultaneous equations by subtracting one equation from the other. This allows us to solve for a first.

    Subtracting Equation 1 from Equation 2 gives us:

    60=(4a+q)(a+q)6=4aa6=3a2=a

    We can substitute the value for a that we have just found into either of the equations. Substituting a into Equation 2 gives us:

    0=12+q2=q

    Therefore the equation for f is

    f(x)=2x2+2

    As your answer you would have entered only the expression on the right hand side:

    2x2+2

    Submit your answer as:

Finding a quadratic equation from its roots

Given that a quadratic equation has the roots -8 and 10, find the equation.

INSTRUCTION: Write the equation in standard form: x2+bx+c=0. (The quadratic coefficient should be 1.)
Answer: Write the entire equation here: .
one-of
type(equation.strict)
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

To find the equation, you should start with the basic form here: (xroot1)(xroot2).


STEP: Find the factors for the equation
[−2 points ⇒ 1 / 3 points left]

To solve this equation, it is like we are solving a quadratic equation backwards: we have the roots of an equation, and now we want to find that equation. Remember that roots are values that make the quadratic expression equal to zero.

Let's take the two super small equations x=8 and x=10 and then subtract the roots over to the left side to get zeroes (be careful with the signs... in this case there are some negatives which become addition):

x=8x(8)=0x+8=0
x=10x(10)=0x10=0

STEP: Find the equation
[−1 point ⇒ 0 / 3 points left]

We have two expressions above ((x+8) and (x10)) which are both equal to zero; if we multiply them, the answer must be zero. We can multiply the two expressions to get the equation.

0=(x+8)(x10)0=x22x80
TIP:

We can check the answer to make sure that it is correct. Remember that the question tells us that the equation we want has a root equal to 8. Therefore, if x=8, the equation must be equal to zero. If x=8:

x22x80(8)22(8)80=0

You can try the same with the second root, and you will find the answer comes out to be zero again!

The correct answer is x22x80=0.


Submit your answer as:

Finding a quadratic equation from its roots

Given that a quadratic equation has the roots 0 and -1, find the equation.

INSTRUCTION: Write the equation in standard form: x2+bx+c=0. (The quadratic coefficient should be 1.)
Answer: Write the entire equation here: .
one-of
type(equation.strict)
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

To find the equation, you should start with the basic form here: (xroot1)(xroot2).


STEP: Find the factors for the equation
[−2 points ⇒ 1 / 3 points left]

To solve this equation, it is like we are solving a quadratic equation backwards: we have the roots of an equation, and now we want to find that equation. Remember that roots are values that make the quadratic expression equal to zero.

Let's take the two super small equations x=0 and x=1 and then subtract the roots over to the left side to get zeroes (be careful with the signs... in this case there are some negatives which become addition):

x=0x(0)=0x=0
x=1x(−1)=0x+1=0

STEP: Find the equation
[−1 point ⇒ 0 / 3 points left]

We have two expressions above ((x) and (x+1)) which are both equal to zero; if we multiply them, the answer must be zero. We can multiply the two expressions to get the equation.

0=(x)(x+1)0=x2+x
TIP:

We can check the answer to make sure that it is correct. Remember that the question tells us that the equation we want has a root equal to 0. Therefore, if x=0, the equation must be equal to zero. If x=0:

x2+x(0)2+(0)=0

You can try the same with the second root, and you will find the answer comes out to be zero again!

The correct answer is x2+x=0.


Submit your answer as:

Finding a quadratic equation from its roots

Find an equation with roots 8 and 8.

INSTRUCTION: Write the equation in standard form: x2+bx+c=0. (The quadratic coefficient should be 1.)
Answer: Write the entire equation here: .
one-of
type(equation.strict)
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

To find the equation, you should start with the basic form here: (xroot1)(xroot2).


STEP: Find the factors for the equation
[−2 points ⇒ 1 / 3 points left]

To solve this equation, it is like we are solving a quadratic equation backwards: we have the roots of an equation, and now we want to find that equation. Remember that roots are values that make the quadratic expression equal to zero.

Let's take the two super small equations x=8 and x=8 and then subtract the roots over to the left side to get zeroes (be careful with the signs... in this case there are some negatives which become addition):

x=8x(8)=0x8=0
x=8x(−8)=0x+8=0

STEP: Find the equation
[−1 point ⇒ 0 / 3 points left]

We have two expressions above ((x8) and (x+8)) which are both equal to zero; if we multiply them, the answer must be zero. We can multiply the two expressions to get the equation.

0=(x8)(x+8)0=x264
TIP:

We can check the answer to make sure that it is correct. Remember that the question tells us that the equation we want has a root equal to 8. Therefore, if x=8, the equation must be equal to zero. If x=8:

x264(8)264=0

You can try the same with the second root, and you will find the answer comes out to be zero again!

The correct answer is x264=0.


Submit your answer as:

4. Quadratic graphs

Calculating the x-intercepts for a quadratic equation

Consider the following equation:

y(x)=9(x1)29

Determine the x-coordinates of the x-intercepts for the function. Give exact answers (do not round off).

INSTRUCTION: Give only the x-coordinate for the intercept points. In other words, each answer should be a number, not a coordinate pair. It does not matter which value you type first.
Answer: The x-coordinates of the x-intercepts are: and .
numeric
numeric
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

First set y=0 in the equation. The solve for x.


STEP: Set y equal to zero
[−1 point ⇒ 3 / 4 points left]

We calculate the x-intercepts by letting y=0. (This means y(x) should change to zero.)

y(x)=9(x1)290=9(x1)29

STEP: Solve for x
[−3 points ⇒ 0 / 4 points left]

Now we need to solve for x. We can do this using inverses. It is important to recognise that there should be two solutions because the equation is quadratic. (It is also possible to solve the equation via factorisation or the quadratic formula. But to do that one must expand the binomial and arrange the equation in standard form. Solving via inverses is faster.)

Start by moving the constant term over to the left side of the equation. The goal is to isolate the exponential expression (x1)2.

0=9(x1)299=9(x1)21=(x1)2

Now take the square root on both sides. We must remember that when we square root both sides of an equation we must bring in a plus-minus. This is because if we use a square root we are undoing a square, and in such a case the number we want could be positive or negative.

cancels the squareThe square root±1=(x1)2±1=x1±1=x1

This means we have two versions of the equation now: one with the positive root, 1 and one with the negative root, 1. We can separate these values into two separate equations and solve each one.

rootpositiverootnegativex1=1x1=1x=1+1x=1+1x=2x=0

This gives us the x-intercepts at these coordinates: (2;0) and (0;0).

NOTE: It is not required for the question, but it is helpful to see the graph of the function. We can see that the intercepts are in the locations that we calculated.

The correct answers are 0 and 2.


Submit your answer as: and

Parabolas: turning points

The figure below shows the graph of the function f(x)=2x24. Answer the two questions which follow about this function.

  1. What are the coordinates of the turning point for the function?

    INSTRUCTION: Type your answer as a coordinate pair, including brackets. Use a semi-colon between the coordinates. For example, your answer might look like this: (3;-1)
    Answer: The turning point is at: .
    coordinate
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Find the turning point on the graph and then figure out its coordinates.


    STEP: Find the turning point
    [−1 point ⇒ 1 / 2 points left]

    We need to find the turning point of the parabola in the graph. Every parabola has a turning point.


    STEP: Write down the coordinates of the turning point
    [−1 point ⇒ 0 / 2 points left]

    Now we just need to read off the coordinates of that turning point.

    NOTE: Remember that the constant term, 4, shifts a graph down 4 spaces. The turning point for the simplest parabola starts at (0;0). So we can find the turning point directly from the function (without the graph). The turning point for the function f(x)=2x24 must be at (0;4).

    The coordinates of the turning point are (0;4).


    Submit your answer as:
  2. What is the minimum value of the function?

    Answer: The minimum value is f= .
    numeric
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    The minimum value of the function refers to the smallest output value. You can find this value at the turning point.


    STEP: Use the turning point to find the minimum value
    [−1 point ⇒ 0 / 1 points left]

    This question asks us to find the minimum value of the function. This means we need to find the lowest value on the parabola. That is the turning point!

    To find the maximum or minimum value of a function, we need to find the maximum or minimum output value for the function. For a parabola, this must be at the turning point. In this case, the turning point shows us a minimum value of 4 because all the other output values are above f=4.

    NOTE: The range of a parabola is determined by the output value at the turning point. This is because the turning point is the limit of the output values. The range for f(x)=2x24 is all y-values greater than or equal to 4.

    The minimum value of the function is f=4.


    Submit your answer as:

The graph of f(x)=x2

The figure below shows the graphs for three different equations. Answer the two questions which follow about these graphs.

  1. Which of the three graphs shows the function f(x)=x2?

    Answer: f(x)=x2 is graph .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    It is important for you to know the shape of the function f(x)=x2. The function f(x)=x2 must go through the point (0;0) because f(0)=0. Which of the graphs go through that point?


    STEP: Find the correct graph
    [−1 point ⇒ 0 / 1 points left]

    The graph of f(x)=x2 has a shape which is important to know. Based on its shape, one can determine a lot of other (more complicated) graphs. As always, the graph of an equation is the set of coordinates which solve the equation. In other words, the graph is made of the x- and y-values which fit in the equation. Here is a table of values for this function:

    x f(x)
    3 9
    2 4
    1 1
    0 0
    1 1
    2 4
    3 9

    These values sit on the graph below:

    The function f(x)=x2 is graph A.


    Submit your answer as:
  2. What is the name of the graph's shape?

    Answer: The graph is called .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    The name is used for the shape of any quadratic equation's graph. If you cannot remember it, check in your notes or a textbook.


    STEP: Select the correct name
    [−1 point ⇒ 0 / 1 points left]

    Any graph which comes from a quadratic equation is called a parabola. Every parabola has a similar shape to the graph in Question 1. Important features of a parabola are:

    • Parabolas always have a turning point. This is different to linear functions, which do not turn around. The turning point comes from the square in the equation.
    • Parabolas are symmetric. A parabola is made of two sides which are reflections of each other. We can see this symmetry in the table in Question 1, which shows that the output values repeat. This symmetry comes from the square in the equation.

    The graph is called a parabola.


    Submit your answer as:

Quadratic functions

The following table shows input and output values for the function f(x)=x2. Note that one of the output values is missing.

x f(x)
3 9
2 4
1 1
0 ?
1 1
2 4
3 9
  1. What is the missing value?
  2. Why are there no negative output values?
Answer:
  1. The missing value is: .
  2. There are no negative output values because .
numeric
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Start by figuring out which x-value to use to find the missing value.


STEP: Calculate the value of f(0)
[−1 point ⇒ 1 / 2 points left]

We need to find the missing value from the table. The missing value is the output value for x=0. So we just need to evaluate the function for x=0.

f(x)=x2f(0)=(0)2=0

The value of f(0) is 0.


STEP: Determine why the output values cannot be negative
[−1 point ⇒ 0 / 2 points left]

We need to figure out why the output values in the table are never negative.

This is a result of the exponent, 2. Whenever we square a number, the answer has to be positive (or zero) - it cannot be negative. This is because if the exponent is 2 and the base is negative, there will be two negative signs. For example:

(9)2negative times a negative!the square leads to a =(9)(9)=81

This is also the reason why the graph of f(x)=x2 "turns around". It cannot go down forever because the output values will start going back up again. So it is the exponent of 2 which gives this graph its special shape.

The correct answers are:

  1. The missing value is 0.
  2. The correct choice is the square cancels negative signs on input values.

Submit your answer as: and

Exercises

Calculating the x-intercepts for a quadratic equation

Consider the following equation:

y(x)=(x1)2+9

Determine the x-coordinates of the x-intercepts for the function. Give exact answers (do not round off).

INSTRUCTION: Give only the x-coordinate for the intercept points. In other words, each answer should be a number, not a coordinate pair. It does not matter which value you type first.
Answer: The x-coordinates of the x-intercepts are: and .
numeric
numeric
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

First set y=0 in the equation. The solve for x.


STEP: Set y equal to zero
[−1 point ⇒ 3 / 4 points left]

We calculate the x-intercepts by letting y=0. (This means y(x) should change to zero.)

y(x)=(x1)2+90=(x1)2+9

STEP: Solve for x
[−3 points ⇒ 0 / 4 points left]

Now we need to solve for x. We can do this using inverses. It is important to recognise that there should be two solutions because the equation is quadratic. (It is also possible to solve the equation via factorisation or the quadratic formula. But to do that one must expand the binomial and arrange the equation in standard form. Solving via inverses is faster.)

Start by moving the constant term over to the left side of the equation. The goal is to isolate the exponential expression (x1)2.

0=(x1)2+99=(x1)29=(x1)2

Now take the square root on both sides. We must remember that when we square root both sides of an equation we must bring in a plus-minus. This is because if we use a square root we are undoing a square, and in such a case the number we want could be positive or negative.

cancels the squareThe square root±9=(x1)2±9=x1±3=x1

This means we have two versions of the equation now: one with the positive root, 3 and one with the negative root, 3. We can separate these values into two separate equations and solve each one.

rootpositiverootnegativex1=3x1=3x=3+1x=3+1x=4x=2

This gives us the x-intercepts at these coordinates: (4;0) and (2;0).

NOTE: It is not required for the question, but it is helpful to see the graph of the function. We can see that the intercepts are in the locations that we calculated.

The correct answers are 2 and 4.


Submit your answer as: and

Calculating the x-intercepts for a quadratic equation

Consider the following equation:

y(x)=9(x+1)24

Determine the x-coordinates of the x-intercepts for the function. Give exact answers (do not round off).

INSTRUCTION: Give only the x-coordinate for the intercept points. In other words, each answer should be a number, not a coordinate pair. It does not matter which value you type first.
Answer: The x-coordinates of the x-intercepts are: and .
numeric
numeric
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

First set y=0 in the equation. The solve for x.


STEP: Set y equal to zero
[−1 point ⇒ 3 / 4 points left]

We calculate the x-intercepts by letting y=0. (This means y(x) should change to zero.)

y(x)=9(x+1)240=9(x+1)24

STEP: Solve for x
[−3 points ⇒ 0 / 4 points left]

Now we need to solve for x. We can do this using inverses. It is important to recognise that there should be two solutions because the equation is quadratic. (It is also possible to solve the equation via factorisation or the quadratic formula. But to do that one must expand the binomial and arrange the equation in standard form. Solving via inverses is faster.)

Start by moving the constant term over to the left side of the equation. The goal is to isolate the exponential expression (x+1)2.

0=9(x+1)244=9(x+1)249=(x+1)2

Now take the square root on both sides. We must remember that when we square root both sides of an equation we must bring in a plus-minus. This is because if we use a square root we are undoing a square, and in such a case the number we want could be positive or negative.

cancels the squareThe square root±49=(x+1)2±49=x+1±23=x+1

This means we have two versions of the equation now: one with the positive root, 23 and one with the negative root, 23. We can separate these values into two separate equations and solve each one.

rootpositiverootnegativex+1=23x+1=23x=231x=231x=13x=53

This gives us the x-intercepts at these coordinates: (13;0) and (53;0).

NOTE: It is not required for the question, but it is helpful to see the graph of the function. We can see that the intercepts are in the locations that we calculated.

The correct answers are 53 and 13.


Submit your answer as: and

Calculating the x-intercepts for a quadratic equation

Consider the following equation:

y(x)=(x+4)24

Determine the x-coordinates of the x-intercepts for the function. Give exact answers (do not round off).

INSTRUCTION: Give only the x-coordinate for the intercept points. In other words, each answer should be a number, not a coordinate pair. It does not matter which value you type first.
Answer: The x-coordinates of the x-intercepts are: and .
numeric
numeric
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

First set y=0 in the equation. The solve for x.


STEP: Set y equal to zero
[−1 point ⇒ 3 / 4 points left]

We calculate the x-intercepts by letting y=0. (This means y(x) should change to zero.)

y(x)=(x+4)240=(x+4)24

STEP: Solve for x
[−3 points ⇒ 0 / 4 points left]

Now we need to solve for x. We can do this using inverses. It is important to recognise that there should be two solutions because the equation is quadratic. (It is also possible to solve the equation via factorisation or the quadratic formula. But to do that one must expand the binomial and arrange the equation in standard form. Solving via inverses is faster.)

Start by moving the constant term over to the left side of the equation. The goal is to isolate the exponential expression (x+4)2.

0=(x+4)244=(x+4)2

Now take the square root on both sides. We must remember that when we square root both sides of an equation we must bring in a plus-minus. This is because if we use a square root we are undoing a square, and in such a case the number we want could be positive or negative.

cancels the squareThe square root±4=(x+4)2±4=x+4±2=x+4

This means we have two versions of the equation now: one with the positive root, 2 and one with the negative root, 2. We can separate these values into two separate equations and solve each one.

rootpositiverootnegativex+4=2x+4=2x=24x=24x=2x=6

This gives us the x-intercepts at these coordinates: (2;0) and (6;0).

NOTE: It is not required for the question, but it is helpful to see the graph of the function. We can see that the intercepts are in the locations that we calculated.

The correct answers are 6 and 2.


Submit your answer as: and

Parabolas: turning points

The figure below shows the graph of the function f(x)=2x21. Answer the two questions which follow about this function.

  1. What are the coordinates of the turning point for the function?

    INSTRUCTION: Type your answer as a coordinate pair, including brackets. Use a semi-colon between the coordinates. For example, your answer might look like this: (3;-1)
    Answer: The turning point is at: .
    coordinate
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Find the turning point on the graph and then figure out its coordinates.


    STEP: Find the turning point
    [−1 point ⇒ 1 / 2 points left]

    We need to find the turning point of the parabola in the graph. Every parabola has a turning point.


    STEP: Write down the coordinates of the turning point
    [−1 point ⇒ 0 / 2 points left]

    Now we just need to read off the coordinates of that turning point.

    NOTE: Remember that the constant term, 1, shifts a graph down 1 space. The turning point for the simplest parabola starts at (0;0). So we can find the turning point directly from the function (without the graph). The turning point for the function f(x)=2x21 must be at (0;1).

    The coordinates of the turning point are (0;1).


    Submit your answer as:
  2. What is the maximum value of the function?

    Answer: The maximum value is f= .
    numeric
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    The maximum value of the function refers to the largest output value. You can find this value at the turning point.


    STEP: Use the turning point to find the maximum value
    [−1 point ⇒ 0 / 1 points left]

    This question asks us to find the maximum value of the function. This means we need to find the highest value on the parabola. That is the turning point!

    To find the maximum or minimum value of a function, we need to find the maximum or minimum output value for the function. For a parabola, this must be at the turning point. In this case, the turning point shows us a maximum value of 1 because all the other output values are less than f=1.

    NOTE: The range of a parabola is determined by the output value at the turning point. This is because the turning point is the limit of the output values. The range for f(x)=2x21 is all y-values less than or equal to 1.

    The maximum value of the function is f=1.


    Submit your answer as:

Parabolas: turning points

The figure below shows the graph of the function f(x)=32x23. Answer the two questions which follow about this function.

  1. What are the coordinates of the turning point for the function?

    INSTRUCTION: Type your answer as a coordinate pair, including brackets. Use a semi-colon between the coordinates. For example, your answer might look like this: (3;-1)
    Answer: The turning point is at: .
    coordinate
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Find the turning point on the graph and then figure out its coordinates.


    STEP: Find the turning point
    [−1 point ⇒ 1 / 2 points left]

    We need to find the turning point of the parabola in the graph. Every parabola has a turning point.


    STEP: Write down the coordinates of the turning point
    [−1 point ⇒ 0 / 2 points left]

    Now we just need to read off the coordinates of that turning point.

    NOTE: Remember that the constant term, 3, shifts a graph down 3 spaces. The turning point for the simplest parabola starts at (0;0). So we can find the turning point directly from the function (without the graph). The turning point for the function f(x)=32x23 must be at (0;3).

    The coordinates of the turning point are (0;3).


    Submit your answer as:
  2. What is the minimum value of the function?

    Answer: The minimum value is f= .
    numeric
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    The minimum value of the function refers to the smallest output value. You can find this value at the turning point.


    STEP: Use the turning point to find the minimum value
    [−1 point ⇒ 0 / 1 points left]

    This question asks us to find the minimum value of the function. This means we need to find the lowest value on the parabola. That is the turning point!

    To find the maximum or minimum value of a function, we need to find the maximum or minimum output value for the function. For a parabola, this must be at the turning point. In this case, the turning point shows us a minimum value of 3 because all the other output values are above f=3.

    NOTE: The range of a parabola is determined by the output value at the turning point. This is because the turning point is the limit of the output values. The range for f(x)=32x23 is all y-values greater than or equal to 3.

    The minimum value of the function is f=3.


    Submit your answer as:

Parabolas: turning points

The figure below shows the graph of the function f(x)=2x23. Answer the two questions which follow about this function.

  1. What are the coordinates of the turning point for the function?

    INSTRUCTION: Type your answer as a coordinate pair, including brackets. Use a semi-colon between the coordinates. For example, your answer might look like this: (3;-1)
    Answer: The turning point is at: .
    coordinate
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Find the turning point on the graph and then figure out its coordinates.


    STEP: Find the turning point
    [−1 point ⇒ 1 / 2 points left]

    We need to find the turning point of the parabola in the graph. Every parabola has a turning point.


    STEP: Write down the coordinates of the turning point
    [−1 point ⇒ 0 / 2 points left]

    Now we just need to read off the coordinates of that turning point.

    NOTE: Remember that the constant term, 3, shifts a graph down 3 spaces. The turning point for the simplest parabola starts at (0;0). So we can find the turning point directly from the function (without the graph). The turning point for the function f(x)=2x23 must be at (0;3).

    The coordinates of the turning point are (0;3).


    Submit your answer as:
  2. What is the maximum value of the function?

    Answer: The maximum value is f= .
    numeric
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    The maximum value of the function refers to the largest output value. You can find this value at the turning point.


    STEP: Use the turning point to find the maximum value
    [−1 point ⇒ 0 / 1 points left]

    This question asks us to find the maximum value of the function. This means we need to find the highest value on the parabola. That is the turning point!

    To find the maximum or minimum value of a function, we need to find the maximum or minimum output value for the function. For a parabola, this must be at the turning point. In this case, the turning point shows us a maximum value of 3 because all the other output values are less than f=3.

    NOTE: The range of a parabola is determined by the output value at the turning point. This is because the turning point is the limit of the output values. The range for f(x)=2x23 is all y-values less than or equal to 3.

    The maximum value of the function is f=3.


    Submit your answer as:

The graph of f(x)=x2

The figure below shows the graphs for three different equations. Answer the two questions which follow about these graphs.

  1. Which of the three graphs shows the function f(x)=x2?

    Answer: f(x)=x2 is graph .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    It is important for you to know the shape of the function f(x)=x2. The function f(x)=x2 must go through the point (0;0) because f(0)=0. Which of the graphs go through that point?


    STEP: Find the correct graph
    [−1 point ⇒ 0 / 1 points left]

    The graph of f(x)=x2 has a shape which is important to know. Based on its shape, one can determine a lot of other (more complicated) graphs. As always, the graph of an equation is the set of coordinates which solve the equation. In other words, the graph is made of the x- and y-values which fit in the equation. Here is a table of values for this function:

    x f(x)
    3 9
    2 4
    1 1
    0 0
    1 1
    2 4
    3 9

    These values sit on the graph below:

    The function f(x)=x2 is graph A.


    Submit your answer as:
  2. What is the name of the graph's shape?

    Answer: The graph is called .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    The name is used for the shape of any quadratic equation's graph. If you cannot remember it, check in your notes or a textbook.


    STEP: Select the correct name
    [−1 point ⇒ 0 / 1 points left]

    Any graph which comes from a quadratic equation is called a parabola. Every parabola has a similar shape to the graph in Question 1. Important features of a parabola are:

    • Parabolas always have a turning point. This is different to linear functions, which do not turn around. The turning point comes from the square in the equation.
    • Parabolas are symmetric. A parabola is made of two sides which are reflections of each other. We can see this symmetry in the table in Question 1, which shows that the output values repeat. This symmetry comes from the square in the equation.

    The graph is called a parabola.


    Submit your answer as:

The graph of f(x)=x2

The figure below shows the graphs for three different equations. Answer the two questions which follow about these graphs.

  1. Which of the three graphs shows the function f(x)=x2?

    Answer: f(x)=x2 is graph .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    It is important for you to know the shape of the function f(x)=x2. The function f(x)=x2 must go through the point (0;0) because f(0)=0. Which of the graphs go through that point?


    STEP: Find the correct graph
    [−1 point ⇒ 0 / 1 points left]

    The graph of f(x)=x2 has a shape which is important to know. Based on its shape, one can determine a lot of other (more complicated) graphs. As always, the graph of an equation is the set of coordinates which solve the equation. In other words, the graph is made of the x- and y-values which fit in the equation. Here is a table of values for this function:

    x f(x)
    3 9
    2 4
    1 1
    0 0
    1 1
    2 4
    3 9

    These values sit on the graph below:

    The function f(x)=x2 is graph C.


    Submit your answer as:
  2. What is the name of the graph's shape?

    Answer: The graph is called .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    The name is used for the shape of any quadratic equation's graph. If you cannot remember it, check in your notes or a textbook.


    STEP: Select the correct name
    [−1 point ⇒ 0 / 1 points left]

    Any graph which comes from a quadratic equation is called a parabola. Every parabola has a similar shape to the graph in Question 1. Important features of a parabola are:

    • Parabolas always have a turning point. This is different to linear functions, which do not turn around. The turning point comes from the square in the equation.
    • Parabolas are symmetric. A parabola is made of two sides which are reflections of each other. We can see this symmetry in the table in Question 1, which shows that the output values repeat. This symmetry comes from the square in the equation.

    The graph is called a parabola.


    Submit your answer as:

The graph of f(x)=x2

The figure below shows the graphs for three different equations. Answer the two questions which follow about these graphs.

  1. Which of the three graphs shows the function f(x)=x2?

    Answer: f(x)=x2 is graph .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    It is important for you to know the shape of the function f(x)=x2. The function f(x)=x2 must go through the point (0;0) because f(0)=0. Which of the graphs go through that point?


    STEP: Find the correct graph
    [−1 point ⇒ 0 / 1 points left]

    The graph of f(x)=x2 has a shape which is important to know. Based on its shape, one can determine a lot of other (more complicated) graphs. As always, the graph of an equation is the set of coordinates which solve the equation. In other words, the graph is made of the x- and y-values which fit in the equation. Here is a table of values for this function:

    x f(x)
    3 9
    2 4
    1 1
    0 0
    1 1
    2 4
    3 9

    These values sit on the graph below:

    The function f(x)=x2 is graph C.


    Submit your answer as:
  2. What is the name of the graph's shape?

    Answer: The graph is called .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    The name is used for the shape of any quadratic equation's graph. If you cannot remember it, check in your notes or a textbook.


    STEP: Select the correct name
    [−1 point ⇒ 0 / 1 points left]

    Any graph which comes from a quadratic equation is called a parabola. Every parabola has a similar shape to the graph in Question 1. Important features of a parabola are:

    • Parabolas always have a turning point. This is different to linear functions, which do not turn around. The turning point comes from the square in the equation.
    • Parabolas are symmetric. A parabola is made of two sides which are reflections of each other. We can see this symmetry in the table in Question 1, which shows that the output values repeat. This symmetry comes from the square in the equation.

    The graph is called a parabola.


    Submit your answer as:

Quadratic functions

The following table shows input and output values for the function f(x)=x2. Note that one of the output values is missing.

x f(x)
3 ?
2 4
1 1
0 0
1 1
2 4
3 9
  1. What is the missing value?
  2. Why are there no negative output values?
Answer:
  1. The missing value is: .
  2. There are no negative output values because .
numeric
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Start by figuring out which x-value to use to find the missing value.


STEP: Calculate the value of f(3)
[−1 point ⇒ 1 / 2 points left]

We need to find the missing value from the table. The missing value is the output value for x=3. So we just need to evaluate the function for x=3.

f(x)=x2f(3)=(3)2=9

The value of f(3) is 9.


STEP: Determine why the output values cannot be negative
[−1 point ⇒ 0 / 2 points left]

We need to figure out why the output values in the table are never negative.

This is a result of the exponent, 2. Whenever we square a number, the answer has to be positive (or zero) - it cannot be negative. This is because if the exponent is 2 and the base is negative, there will be two negative signs. For example:

(4)2negative times a negative!the square leads to a =(4)(4)=16

This is also the reason why the graph of f(x)=x2 "turns around". It cannot go down forever because the output values will start going back up again. So it is the exponent of 2 which gives this graph its special shape.

The correct answers are:

  1. The missing value is 9.
  2. The correct choice is the square cancels negative signs on input values.

Submit your answer as: and

Quadratic functions

The following table shows input and output values for the function f(x)=x2. Note that one of the output values is missing.

x f(x)
3 9
2 ?
1 1
0 0
1 1
2 4
3 9
  1. What is the missing value?
  2. Why are there no negative output values?
Answer:
  1. The missing value is: .
  2. There are no negative output values because .
numeric
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Start by figuring out which x-value to use to find the missing value.


STEP: Calculate the value of f(2)
[−1 point ⇒ 1 / 2 points left]

We need to find the missing value from the table. The missing value is the output value for x=2. So we just need to evaluate the function for x=2.

f(x)=x2f(2)=(2)2=4

The value of f(2) is 4.


STEP: Determine why the output values cannot be negative
[−1 point ⇒ 0 / 2 points left]

We need to figure out why the output values in the table are never negative.

This is a result of the exponent, 2. Whenever we square a number, the answer has to be positive (or zero) - it cannot be negative. This is because if the exponent is 2 and the base is negative, there will be two negative signs. For example:

(8)2negative times a negative!the square leads to a =(8)(8)=64

This is also the reason why the graph of f(x)=x2 "turns around". It cannot go down forever because the output values will start going back up again. So it is the exponent of 2 which gives this graph its special shape.

The correct answers are:

  1. The missing value is 4.
  2. The correct choice is the square cancels negative signs on input values.

Submit your answer as: and

Quadratic functions

The following table shows input and output values for the function f(x)=x2. Note that one of the output values is missing.

x f(x)
3 9
2 4
1 ?
0 0
1 1
2 4
3 9
  1. What is the missing value?
  2. Why are there no negative output values?
Answer:
  1. The missing value is: .
  2. There are no negative output values because .
numeric
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Start by figuring out which x-value to use to find the missing value.


STEP: Calculate the value of f(1)
[−1 point ⇒ 1 / 2 points left]

We need to find the missing value from the table. The missing value is the output value for x=1. So we just need to evaluate the function for x=1.

f(x)=x2f(1)=(1)2=1

The value of f(1) is 1.


STEP: Determine why the output values cannot be negative
[−1 point ⇒ 0 / 2 points left]

We need to figure out why the output values in the table are never negative.

This is a result of the exponent, 2. Whenever we square a number, the answer has to be positive (or zero) - it cannot be negative. This is because if the exponent is 2 and the base is negative, there will be two negative signs. For example:

(8)2negative times a negative!the square leads to a =(8)(8)=64

This is also the reason why the graph of f(x)=x2 "turns around". It cannot go down forever because the output values will start going back up again. So it is the exponent of 2 which gives this graph its special shape.

The correct answers are:

  1. The missing value is 1.
  2. The correct choice is the square cancels negative signs on input values.

Submit your answer as: and

5. Completing the square

Completing the square: when a1

  1. Suppose we start with the expression shown here:

    2a2+2a

    If we want to complete the square for this expression, we must add the correct number. We also must subtract that same number so that the overall value of the expression does not change. If N represents the correct number to add and subtract, we are looking for something like this:

    2a2+2a+NN

    Determine the value of N, which will complete the square. Your answer must be exact (not rounded).

    Answer: The number to complete the square is .
    numeric
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    The quadratic coefficient of 2 is quite a nuisance in this question. The only way to get it out of the way is to factorise it out of the original expression.


    STEP: Factorise out the quadratic coefficient, 2
    [−1 point ⇒ 1 / 2 points left]

    If we must complete the square for an expression and the quadratic coefficient is not 1, we need to change the expression so that the quadratic coefficient is "out of the way." That means we must factorise the quadratic coefficient out of the expression. This may bring fractions into the expression, but there is no other way!

    2a2+2a+NN=2(a2+a+N2N2)

    Now complete the square inside the brackets by using the value (b2)2. It is always a good idea when completing the square to calculate the value of b2, which you will probably use later. b2=12b=12(1)=12. Then, the number we must add (and subtract) inside the brackets to complete the square is:

    (b2)2=(12)2=14

    Therefore, we must add 14 to the expression.


    STEP: Determine the value of N
    [−1 point ⇒ 0 / 2 points left]

    We need to compare the value we calculated above to the value of N. We found above that we need to add (and subtract) 14 to the expression a2+a, which is the expression inside of the brackets. So:

    2(a2+a+N2N2)must become:2(a2+a+1414)

    So that means:

    N2=14N=2(14)N=12
    NOTE:

    You can see that the number we get is actually 2(14). This is the number we must add (and subtract) to the original expression. The factor of 2 comes from the quadratic coefficient, which we factorised out. But it is still part of the expression. The original expression, with the value of N included, is:

    2a2+2a+1212

    The final answer is: N=12.


    Submit your answer as:
  2. Now factorise the completed expression from above so that it contains a squared binomial. For example, it might look like: 2(x259)214.

    Answer: The expression will be: .
    expression
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    You need to factorise the expression 2(a2+a+1414) so that it has the arrangement 2(a±?)2?.


    STEP: Use the result from Question 1 to get the answer
    [−2 points ⇒ 0 / 2 points left]

    The first thing we need to do is separate out the last term from the other terms. We do this because we can only factorise the other terms (the last term is only there as an echo from our work to complete the square on the first two terms).

    2(a2+a+1414)=2(a2+a+14)12

    Now we can factorise the trinomial in the brackets:

    2(a2+a+14)12=2(a+12)(a+12)12=2(a+12)212

    The correct answer is 2(a+12)212.


    Submit your answer as:

Completing the square

  1. For the expression shown here, determine what value (what number) will complete the square.

    k2+16k+?
    Answer:

    The number to complete the square is .

    numeric
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    You need to use the linear coefficient, 16, to find the answer. To complete the square, you must add (b2)2 to the expression, where b is the linear coefficient.


    STEP: Use the linear coefficient to figure out the number which will complete the square
    [−1 point ⇒ 0 / 1 points left]

    To complete the square, we need to add a number to the expression so that it will fit the squared binomial pattern

    (m+n)2=m2+2mn+n2

    If b is the linear coefficient, then we must add the value (b2)2 to complete the square. In the expression above, b is equal to 16. Start with the value

    b2=(16)2=8

    This value is almost always important when you complete the square. Then, the number we must add to complete the square is:

    (b2)2=(8)2=64

    When we complete the square, the expression becomes k2+16k+64 (you will use this in the next question).

    The number we need to complete the square is 64.


    Submit your answer as:
  2. Now factorise the completed expression from above so that it is a squared binomial. For example, the answer might look like (x3)2.

    Answer: The expression will be: .
    expression
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    You need to factorise the expression k2+16k+64 so that it has the arrangement (k±?)2.


    STEP: Use the answer from Question 1 to write down the answer
    [−1 point ⇒ 0 / 1 points left]

    In the first part of the problem, we found that b2=8; this is the value which sits in the binomial when we factorise:

    k2+16k+64=(k+8)(k+8)=(k+8)2

    It is clear now why the value for completing the square will be the square of 8: when we multiply the brackets (with FOIL) one of the terms we get will be 8 multiplied by itself - that is why completing the square uses the number (b2)2.

    The correct answer is (k+8)2.


    Submit your answer as:

Exercises

Completing the square: when a1

  1. Suppose we start with the expression shown here:

    2d2+2d

    If we want to complete the square for this expression, we must add the correct number. We also must subtract that same number so that the overall value of the expression does not change. If N represents the correct number to add and subtract, we are looking for something like this:

    2d2+2d+NN

    Determine the value of N, which will complete the square. Your answer must be exact (not rounded).

    Answer: The number to complete the square is .
    numeric
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    The quadratic coefficient of 2 is quite a nuisance in this question. The only way to get it out of the way is to factorise it out of the original expression.


    STEP: Factorise out the quadratic coefficient, 2
    [−1 point ⇒ 1 / 2 points left]

    If we must complete the square for an expression and the quadratic coefficient is not 1, we need to change the expression so that the quadratic coefficient is "out of the way." That means we must factorise the quadratic coefficient out of the expression. This may bring fractions into the expression, but there is no other way!

    2d2+2d+NN=2(d2+d+N2N2)

    Now complete the square inside the brackets by using the value (b2)2. It is always a good idea when completing the square to calculate the value of b2, which you will probably use later. b2=12b=12(1)=12. Then, the number we must add (and subtract) inside the brackets to complete the square is:

    (b2)2=(12)2=14

    Therefore, we must add 14 to the expression.


    STEP: Determine the value of N
    [−1 point ⇒ 0 / 2 points left]

    We need to compare the value we calculated above to the value of N. We found above that we need to add (and subtract) 14 to the expression d2+d, which is the expression inside of the brackets. So:

    2(d2+d+N2N2)must become:2(d2+d+1414)

    So that means:

    N2=14N=2(14)N=12
    NOTE:

    You can see that the number we get is actually 2(14). This is the number we must add (and subtract) to the original expression. The factor of 2 comes from the quadratic coefficient, which we factorised out. But it is still part of the expression. The original expression, with the value of N included, is:

    2d2+2d+1212

    The final answer is: N=12.


    Submit your answer as:
  2. Now factorise the completed expression from above so that it contains a squared binomial. For example, it might look like: 2(x259)214.

    Answer: The expression will be: .
    expression
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    You need to factorise the expression 2(d2+d+1414) so that it has the arrangement 2(d±?)2?.


    STEP: Use the result from Question 1 to get the answer
    [−2 points ⇒ 0 / 2 points left]

    The first thing we need to do is separate out the last term from the other terms. We do this because we can only factorise the other terms (the last term is only there as an echo from our work to complete the square on the first two terms).

    2(d2+d+1414)=2(d2+d+14)12

    Now we can factorise the trinomial in the brackets:

    2(d2+d+14)12=2(d+12)(d+12)12=2(d+12)212

    The correct answer is 2(d+12)212.


    Submit your answer as:

Completing the square: when a1

  1. Suppose we start with the expression shown here:

    3d22d

    If we want to complete the square for this expression, we must add the correct number. We also must subtract that same number so that the overall value of the expression does not change. If N represents the correct number to add and subtract, we are looking for something like this:

    3d22d+NN

    Determine the value of N, which will complete the square. Your answer must be exact (not rounded).

    Answer: The number to complete the square is .
    numeric
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    The quadratic coefficient of 3 is quite a nuisance in this question. The only way to get it out of the way is to factorise it out of the original expression.


    STEP: Factorise out the quadratic coefficient, 3
    [−1 point ⇒ 1 / 2 points left]

    If we must complete the square for an expression and the quadratic coefficient is not 1, we need to change the expression so that the quadratic coefficient is "out of the way." That means we must factorise the quadratic coefficient out of the expression. This may bring fractions into the expression, but there is no other way!

    3d22d+NN=3(d22d3+N3N3)

    Now complete the square inside the brackets by using the value (b2)2. It is always a good idea when completing the square to calculate the value of b2, which you will probably use later. b2=12b=12(23)=13. Then, the number we must add (and subtract) inside the brackets to complete the square is:

    (b2)2=(13)2=19

    Therefore, we must add 19 to the expression.


    STEP: Determine the value of N
    [−1 point ⇒ 0 / 2 points left]

    We need to compare the value we calculated above to the value of N. We found above that we need to add (and subtract) 19 to the expression d22d3, which is the expression inside of the brackets. So:

    3(d22d3+N3N3)must become:3(d22d3+1919)

    So that means:

    N3=19N=3(19)N=13
    NOTE:

    You can see that the number we get is actually 3(19). This is the number we must add (and subtract) to the original expression. The factor of 3 comes from the quadratic coefficient, which we factorised out. But it is still part of the expression. The original expression, with the value of N included, is:

    3d22d+1313

    The final answer is: N=13.


    Submit your answer as:
  2. Now factorise the completed expression from above so that it contains a squared binomial. For example, it might look like: 2(x259)214.

    Answer: The expression will be: .
    expression
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    You need to factorise the expression 3(d22d3+1919) so that it has the arrangement 3(d±?)2?.


    STEP: Use the result from Question 1 to get the answer
    [−2 points ⇒ 0 / 2 points left]

    The first thing we need to do is separate out the last term from the other terms. We do this because we can only factorise the other terms (the last term is only there as an echo from our work to complete the square on the first two terms).

    3(d22d3+1919)=3(d22d3+19)13

    Now we can factorise the trinomial in the brackets:

    3(d22d3+19)13=3(d13)(d13)13=3(d13)213

    The correct answer is 3(d13)213.


    Submit your answer as:

Completing the square: when a1

  1. Suppose we start with the expression shown here:

    2p2+4p

    If we want to complete the square for this expression, we must add the correct number. We also must subtract that same number so that the overall value of the expression does not change. If N represents the correct number to add and subtract, we are looking for something like this:

    2p2+4p+NN

    Determine the value of N, which will complete the square. Your answer must be exact (not rounded).

    Answer: The number to complete the square is .
    numeric
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    The quadratic coefficient of 2 is quite a nuisance in this question. The only way to get it out of the way is to factorise it out of the original expression.


    STEP: Factorise out the quadratic coefficient, 2
    [−1 point ⇒ 1 / 2 points left]

    If we must complete the square for an expression and the quadratic coefficient is not 1, we need to change the expression so that the quadratic coefficient is "out of the way." That means we must factorise the quadratic coefficient out of the expression. This may bring fractions into the expression, but there is no other way!

    2p2+4p+NN=2(p2+2p+N2N2)

    Now complete the square inside the brackets by using the value (b2)2. It is always a good idea when completing the square to calculate the value of b2, which you will probably use later. b2=12b=12(2)=1. Then, the number we must add (and subtract) inside the brackets to complete the square is:

    (b2)2=(1)2=1

    Therefore, we must add 1 to the expression.


    STEP: Determine the value of N
    [−1 point ⇒ 0 / 2 points left]

    We need to compare the value we calculated above to the value of N. We found above that we need to add (and subtract) 1 to the expression p2+2p, which is the expression inside of the brackets. So:

    2(p2+2p+N2N2)must become:2(p2+2p+11)

    So that means:

    N2=1N=2(1)N=2
    NOTE:

    You can see that the number we get is actually 2(1). This is the number we must add (and subtract) to the original expression. The factor of 2 comes from the quadratic coefficient, which we factorised out. But it is still part of the expression. The original expression, with the value of N included, is:

    2p2+4p+22

    The final answer is: N=2.


    Submit your answer as:
  2. Now factorise the completed expression from above so that it contains a squared binomial. For example, it might look like: 2(x259)214.

    Answer: The expression will be: .
    expression
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    You need to factorise the expression 2(p2+2p+11) so that it has the arrangement 2(p±?)2?.


    STEP: Use the result from Question 1 to get the answer
    [−2 points ⇒ 0 / 2 points left]

    The first thing we need to do is separate out the last term from the other terms. We do this because we can only factorise the other terms (the last term is only there as an echo from our work to complete the square on the first two terms).

    2(p2+2p+11)=2(p2+2p+1)2

    Now we can factorise the trinomial in the brackets:

    2(p2+2p+1)2=2(p+1)(p+1)2=2(p+1)22

    The correct answer is 2(p+1)22.


    Submit your answer as:

Completing the square

  1. For the expression shown here, determine what value (what number) will complete the square.

    d212d+?
    Answer:

    The number to complete the square is .

    numeric
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    You need to use the linear coefficient, 12, to find the answer. To complete the square, you must add (b2)2 to the expression, where b is the linear coefficient.


    STEP: Use the linear coefficient to figure out the number which will complete the square
    [−1 point ⇒ 0 / 1 points left]

    To complete the square, we need to add a number to the expression so that it will fit the squared binomial pattern

    (m+n)2=m2+2mn+n2

    If b is the linear coefficient, then we must add the value (b2)2 to complete the square. In the expression above, b is equal to 12. Start with the value

    b2=(12)2=6

    This value is almost always important when you complete the square. Then, the number we must add to complete the square is:

    (b2)2=(6)2=36

    When we complete the square, the expression becomes d212d+36 (you will use this in the next question).

    The number we need to complete the square is 36.


    Submit your answer as:
  2. Now factorise the completed expression from above so that it is a squared binomial. For example, the answer might look like (x3)2.

    Answer: The expression will be: .
    expression
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    You need to factorise the expression d212d+36 so that it has the arrangement (d±?)2.


    STEP: Use the answer from Question 1 to write down the answer
    [−1 point ⇒ 0 / 1 points left]

    In the first part of the problem, we found that b2=6; this is the value which sits in the binomial when we factorise:

    d212d+36=(d6)(d6)=(d6)2

    It is clear now why the value for completing the square will be the square of 6: when we multiply the brackets (with FOIL) one of the terms we get will be 6 multiplied by itself - that is why completing the square uses the number (b2)2.

    The correct answer is (d6)2.


    Submit your answer as:

Completing the square

  1. For the expression shown here, determine what value (what number) will complete the square.

    d2+12d+?
    Answer:

    The number to complete the square is .

    numeric
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    You need to use the linear coefficient, 12, to find the answer. To complete the square, you must add (b2)2 to the expression, where b is the linear coefficient.


    STEP: Use the linear coefficient to figure out the number which will complete the square
    [−1 point ⇒ 0 / 1 points left]

    To complete the square, we need to add a number to the expression so that it will fit the squared binomial pattern

    (m+n)2=m2+2mn+n2

    If b is the linear coefficient, then we must add the value (b2)2 to complete the square. In the expression above, b is equal to 12. Start with the value

    b2=(12)2=6

    This value is almost always important when you complete the square. Then, the number we must add to complete the square is:

    (b2)2=(6)2=36

    When we complete the square, the expression becomes d2+12d+36 (you will use this in the next question).

    The number we need to complete the square is 36.


    Submit your answer as:
  2. Now factorise the completed expression from above so that it is a squared binomial. For example, the answer might look like (x3)2.

    Answer: The expression will be: .
    expression
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    You need to factorise the expression d2+12d+36 so that it has the arrangement (d±?)2.


    STEP: Use the answer from Question 1 to write down the answer
    [−1 point ⇒ 0 / 1 points left]

    In the first part of the problem, we found that b2=6; this is the value which sits in the binomial when we factorise:

    d2+12d+36=(d+6)(d+6)=(d+6)2

    It is clear now why the value for completing the square will be the square of 6: when we multiply the brackets (with FOIL) one of the terms we get will be 6 multiplied by itself - that is why completing the square uses the number (b2)2.

    The correct answer is (d+6)2.


    Submit your answer as:

Completing the square

  1. For the expression shown here, determine what value (what number) will complete the square.

    a2+2a+?
    Answer:

    The number to complete the square is .

    numeric
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    You need to use the linear coefficient, 2, to find the answer. To complete the square, you must add (b2)2 to the expression, where b is the linear coefficient.


    STEP: Use the linear coefficient to figure out the number which will complete the square
    [−1 point ⇒ 0 / 1 points left]

    To complete the square, we need to add a number to the expression so that it will fit the squared binomial pattern

    (m+n)2=m2+2mn+n2

    If b is the linear coefficient, then we must add the value (b2)2 to complete the square. In the expression above, b is equal to 2. Start with the value

    b2=(2)2=1

    This value is almost always important when you complete the square. Then, the number we must add to complete the square is:

    (b2)2=(1)2=1

    When we complete the square, the expression becomes a2+2a+1 (you will use this in the next question).

    The number we need to complete the square is 1.


    Submit your answer as:
  2. Now factorise the completed expression from above so that it is a squared binomial. For example, the answer might look like (x3)2.

    Answer: The expression will be: .
    expression
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    You need to factorise the expression a2+2a+1 so that it has the arrangement (a±?)2.


    STEP: Use the answer from Question 1 to write down the answer
    [−1 point ⇒ 0 / 1 points left]

    In the first part of the problem, we found that b2=1; this is the value which sits in the binomial when we factorise:

    a2+2a+1=(a+1)(a+1)=(a+1)2

    It is clear now why the value for completing the square will be the square of 1: when we multiply the brackets (with FOIL) one of the terms we get will be 1 multiplied by itself - that is why completing the square uses the number (b2)2.

    The correct answer is (a+1)2.


    Submit your answer as:

6. Quadratic formula

The quadratic formula: solving quadratic equations

Solve for z using the formula.

z22z+5=0
NOTE: While it may be possible to solve this equation by factorisation, the question says that you should solve the equation using the quadratic formula; you will only earn full marks on a test or exam if you solve it using the formula.
INSTRUCTIONS:
  • Round irrational answers to two decimal places.
  • If the answer is non-real, type non-real in the answer box.
  • If there are two answers, separate the answer with a semi-colon.
    For example: 1,23 ; 4,56. The order of the answers does not matter.
Answer: z=
one-of
type(string.nocase)
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

First make sure that the equation is in standard form. Then find the values of a, b and c. Substitute them into the quadratic formula and go for it!


STEP: Determine the values of a,b and c
[−1 point ⇒ 3 / 4 points left]

This equation is already in standard form, so we can read the values of a, b and c directly from the equation:

z22z+5=0
a=1,b=2,c=5

STEP: Use the quadratic formula
[−3 points ⇒ 0 / 4 points left]

Now write down the formula, and substitute in the values for a, b and c. Remember to substitute with brackets, which helps with the negative signs!

z=b±b24ac2a=(2)±(2)24(1)(5)2(1)=2±4202=2±162

Oh snap! We have found the square root of a negative number. There is nothing more for us to do: the answer is 'non-real' for this question.


Submit your answer as:

Quadratic equations: finding the values of a, b and c

The equation here is in standard form for a quadratic equation. What are the values for a, b and c? (Give your answers for the equation that you see: do not alter the equation in any way first.)

0=k2+5k+2
Answer: a= b= c=
numeric
numeric
numeric
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]
Standard form for a quadratic equation looks like this: 0=ax2+bx+c --OR-- ax2+bx+c=0. You just need to read the values of a, b and c from the equation.
STEP: <no title>
[−3 points ⇒ 0 / 3 points left]

In "standard form," the values of a, b and c are as follows: 0=ax2+bx+c --or-- ax2+bx+c=0 a is the coefficient of the quadratic term; b is the coefficient of the linear term; c is the value of the constant term.

To answer this question, you must read off the values from the equation:

0=k2+5k+2a=1b=5c=2


Submit your answer as: andand

The quadratic formula: managing equations first

Solve the equation given with the quadratic formula.

p2+3p2=0
INSTRUCTIONS:
  • List your answers separated by a semi-colon ; .
  • Give your answer(s) as a simplified fraction.
  • If the answer(s) are not real numbers, type non-real.
Answer: p=
list
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

The equation is in standard form, but there is something you should do to help yourself make the question simpler before you use the quadratic formula.


STEP: <no title>
[−1 point ⇒ 3 / 4 points left]

Before starting to solve the equation, it is a good idea to multiply the entire equation by 1 to remove the negative sign from the quadratic term. (This is not a necessary step, but there are numerous reasons why it is helpful in lots of cases!)

p2+3p2=0(1)(p2+3p2)=(0)(1)p23p+2=0

STEP: <no title>
[−3 points ⇒ 0 / 4 points left]

You have the value for a, b and c sitting in the equation now, so write down the formula and go for it. Remember to substitute with brackets, which helps with the negative signs!

a=1,b=3,c=2p=b±b24ac2a=(3)±(3)24(1)(2)2(1)=3±982=3±12

The radicand in this question (the number under the root) is a square number! That means the answers will be rational numbers, because the square root will tidy up very nicely.

It is a good idea at this point to write the two different answer separately.

p=3±12=3+12and =312=42and =22

The final answers are p=2 and p=1.


Submit your answer as:

Quadratic equations: standard form

Given the quadratic equation:

3k=3k22

Which of the following shows the equation in standard form?

Answer:
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Standard form for a quadratic equation looks like this:

0=ax2+bx+c or ax2+bx+c=0

STEP: Rearrange the terms to put the equation in standard form
[−2 points ⇒ 0 / 2 points left]

"Standard form" is a certain arrangement of the equation: there must be a zero on one side and the other terms must be arranged in descending order:

0=ax2+bx+c or ax2+bx+c=0

We can choose if we want to move the terms to the left side of the equation or the right side of the equation. However, it is almost always best to make the quadratic coefficient positive. In this case, that means we should move the terms onto the left side of the equation:

3k=3k223k23k+2=0

In fact there are many ways to arrange the equation in standard form. For example, another possible answer is:

3k2+3k2=0

One correct answer (and probably the best answer!) is: 3k23k+2=0.


Submit your answer as:

Exercises

The quadratic formula: solving quadratic equations

Use the quadratic formula to solve the equation shown.

0=4z2+z+2
NOTE: While it may be possible to solve this equation by factorisation, the question says that you should solve the equation using the quadratic formula; you will only earn full marks on a test or exam if you solve it using the formula.
INSTRUCTIONS:
  • Round irrational answers to two decimal places.
  • If the answer is non-real, type non-real in the answer box.
  • If there are two answers, separate the answer with a semi-colon.
    For example: 1,23 ; 4,56. The order of the answers does not matter.
Answer: z=
one-of
type(string.nocase)
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

First make sure that the equation is in standard form. Then find the values of a, b and c. Substitute them into the quadratic formula and go for it!


STEP: Determine the values of a,b and c
[−1 point ⇒ 3 / 4 points left]

This equation is already in standard form, so we can read the values of a, b and c directly from the equation:

0=4z2+z+2
a=4,b=1,c=2

STEP: Use the quadratic formula
[−3 points ⇒ 0 / 4 points left]

Now write down the formula, and substitute in the values for a, b and c. Remember to substitute with brackets, which helps with the negative signs!

z=b±b24ac2a=(1)±(1)24(4)(2)2(4)=1±1328=1±318

Oh snap! We have found the square root of a negative number. There is nothing more for us to do: the answer is 'non-real' for this question.


Submit your answer as:

The quadratic formula: solving quadratic equations

Use the quadratic formula to solve the equation shown.

3n2+n1=6n24n+1
NOTE: While it may be possible to solve this equation by factorisation, the question says that you should solve the equation using the quadratic formula; you will only earn full marks on a test or exam if you solve it using the formula.
INSTRUCTIONS:
  • Round irrational answers to two decimal places.
  • If the answer is non-real, type non-real in the answer box.
  • If there are two answers, separate the answer with a semi-colon.
    For example: 1,23 ; 4,56. The order of the answers does not matter.
Answer: n=
list
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

First make sure that the equation is in standard form. Then find the values of a, b and c. Substitute them into the quadratic formula and go for it!


STEP: Determine the values of a,b and c
[−2 points ⇒ 3 / 5 points left]

This equation is not in standard form! The first thing we must do is move the terms so that we have standard form. Remember that you should try to make the quadratic coefficient positive when you do this! After that, we can find the values of a, b and c.

3n2+n1=6n24n+10=3n25n+2
a=3,b=5,c=2

STEP: Use the quadratic formula
[−3 points ⇒ 0 / 5 points left]

Now write down the formula, and substitute in the values for a, b and c. Remember to substitute with brackets, which helps with the negative signs!

n=b±b24ac2a=(5)±(5)24(3)(2)2(3)=5±25246=5±16

The radicand in this question (the number under the root) is a square number! That means the answers will be rational numbers, because the square root will tidy up very nicely.

It is a good idea at this point to write the two different answer separately.

n=5±16=5+16and =516=66=46

The final answers are n=1 and n=23.


Submit your answer as:

The quadratic formula: solving quadratic equations

Solve for n using the formula.

n2+3n+3=n1
NOTE: While it may be possible to solve this equation by factorisation, the question says that you should solve the equation using the quadratic formula; you will only earn full marks on a test or exam if you solve it using the formula.
INSTRUCTIONS:
  • Round irrational answers to two decimal places.
  • If the answer is non-real, type non-real in the answer box.
  • If there are two answers, separate the answer with a semi-colon.
    For example: 1,23 ; 4,56. The order of the answers does not matter.
Answer: n=
one-of
type(list.type(numeric.noerror).unordered)
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

First make sure that the equation is in standard form. Then find the values of a, b and c. Substitute them into the quadratic formula and go for it!


STEP: Determine the values of a,b and c
[−2 points ⇒ 3 / 5 points left]

This equation is not in standard form! The first thing we must do is move the terms so that we have standard form. Remember that you should try to make the quadratic coefficient positive when you do this! After that, we can find the values of a, b and c.

n2+3n+3=n1n2+4n+4=0
a=1,b=4,c=4

STEP: Use the quadratic formula
[−3 points ⇒ 0 / 5 points left]

Now write down the formula, and substitute in the values for a, b and c. Remember to substitute with brackets, which helps with the negative signs!

n=b±b24ac2a=(4)±(4)24(1)(4)2(1)=4±16162=4±02

In this case, the radicand (the number in the square root) is zero. That means there will only be one answer.

n=4±02=42=2

The final answer is n=2.


Submit your answer as:

Quadratic equations: finding the values of a, b and c

The equation here is in standard form for a quadratic equation. What are the values for a, b and c? (Give your answers for the equation that you see: do not alter the equation in any way first.)

5y2+2y2=0
Answer: a= b= c=
numeric
numeric
numeric
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]
Standard form for a quadratic equation looks like this: 0=ax2+bx+c --OR-- ax2+bx+c=0. You just need to read the values of a, b and c from the equation.
STEP: <no title>
[−3 points ⇒ 0 / 3 points left]

In "standard form," the values of a, b and c are as follows: 0=ax2+bx+c --or-- ax2+bx+c=0 a is the coefficient of the quadratic term; b is the coefficient of the linear term; c is the value of the constant term.

To answer this question, you must read off the values from the equation:

5y2+2y2=0a=5b=2c=2


Submit your answer as: andand

Quadratic equations: finding the values of a, b and c

The equation here is in standard form for a quadratic equation. What are the values for a, b and c? (Give your answers for the equation that you see: do not alter the equation in any way first.)

4y22=0
Answer: a= b= c=
numeric
numeric
numeric
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]
Standard form for a quadratic equation looks like this: 0=ax2+bx+c --OR-- ax2+bx+c=0. You just need to read the values of a, b and c from the equation.
STEP: <no title>
[−3 points ⇒ 0 / 3 points left]

In "standard form," the values of a, b and c are as follows: 0=ax2+bx+c --or-- ax2+bx+c=0 a is the coefficient of the quadratic term; b is the coefficient of the linear term; c is the value of the constant term.

To answer this question, you must read off the values from the equation:

4y22=0a=4b=0c=2

Notice that in this case, there is no linear term, which means that b must be equal to zero.


Submit your answer as: andand

Quadratic equations: finding the values of a, b and c

The equation here is in standard form for a quadratic equation. What are the values for a, b and c? (Give your answers for the equation that you see: do not alter the equation in any way first.)

0=3p2+5p2
Answer: a= b= c=
numeric
numeric
numeric
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]
Standard form for a quadratic equation looks like this: 0=ax2+bx+c --OR-- ax2+bx+c=0. You just need to read the values of a, b and c from the equation.
STEP: <no title>
[−3 points ⇒ 0 / 3 points left]

In "standard form," the values of a, b and c are as follows: 0=ax2+bx+c --or-- ax2+bx+c=0 a is the coefficient of the quadratic term; b is the coefficient of the linear term; c is the value of the constant term.

To answer this question, you must read off the values from the equation:

0=3p2+5p2a=3b=5c=2


Submit your answer as: andand

The quadratic formula: managing equations first

Use the quadratic formula to solve the equation shown.

0=n23+2n3+13
INSTRUCTIONS:
  • List your answers separated by a semi-colon ; .
  • Give your answer(s) as a decimal rounded to two places if appropriate.
  • If the answer(s) are not real numbers, type non-real.
Answer: n=
numeric
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

The equation is in standard form, but there is something you should do to help yourself make the question simpler before you use the quadratic formula.


STEP: <no title>
[−1 point ⇒ 3 / 4 points left]

Before starting to solve the equation, it is a good idea to multiply the entire equation by 3 because of the fractions in the equation: cancel the denominators before you substitute into the quadratic formula. (This is not a necessary step, but there are numerous reasons why it is helpful in lots of cases!)

0=n23+2n3+13(3)(0)=(n23+2n3+13)(3)0=n2+2n+1

STEP: <no title>
[−3 points ⇒ 0 / 4 points left]

You have the value for a, b and c sitting in the equation now, so write down the formula and go for it. Remember to substitute with brackets, which helps with the negative signs!

a=1,b=2,c=1n=b±b24ac2a=(2)±(2)24(1)(1)2(1)=2±442=2±02

In this case, the radicand (the number in the square root) is zero. That means there will only be one answer.

n=2±02=22=1

The final answer is n=1.


Submit your answer as:

The quadratic formula: managing equations first

Use the quadratic formula to solve the equation shown.

6z22z10=0
INSTRUCTIONS:
  • List your answers separated by a semi-colon ; .
  • Give your answer(s) as a decimal rounded to two places if appropriate.
  • If the answer(s) are not real numbers, type non-real.
Answer: z=
one-of
type(string.nocase)
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

The equation is in standard form, but there is something you should do to help yourself make the question simpler before you use the quadratic formula.


STEP: <no title>
[−1 point ⇒ 3 / 4 points left]

Before starting to solve the equation, it is a good idea to multiply the entire equation by 12 because all of the coefficients in the equation have a common factor of -2. (This is not a necessary step, but there are numerous reasons why it is helpful in lots of cases!)

6z22z10=0(12)(6z22z10)=(0)(12)3z2+z+5=0

STEP: <no title>
[−3 points ⇒ 0 / 4 points left]

You have the value for a, b and c sitting in the equation now, so write down the formula and go for it. Remember to substitute with brackets, which helps with the negative signs!

a=3,b=1,c=5z=b±b24ac2a=(1)±(1)24(3)(5)2(3)=1±1606=1±596

Oh snap! We have found the square root of a negative number. There is nothing more for us to do: the answer is 'non-real' for this question.


Submit your answer as:

The quadratic formula: managing equations first

Solve the equation given with the quadratic formula.

6n2+21n9=0
INSTRUCTIONS:
  • List your answers separated by a semi-colon ; .
  • Give your answer(s) as a simplified fraction.
  • If the answer(s) are not real numbers, type non-real.
Answer: n=
list
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

The equation is in standard form, but there is something you should do to help yourself make the question simpler before you use the quadratic formula.


STEP: <no title>
[−1 point ⇒ 3 / 4 points left]

Before starting to solve the equation, it is a good idea to multiply the entire equation by 13 because all of the coefficients in the equation have a common factor of -3. (This is not a necessary step, but there are numerous reasons why it is helpful in lots of cases!)

6n2+21n9=0(13)(6n2+21n9)=(0)(13)2n27n+3=0

STEP: <no title>
[−3 points ⇒ 0 / 4 points left]

You have the value for a, b and c sitting in the equation now, so write down the formula and go for it. Remember to substitute with brackets, which helps with the negative signs!

a=2,b=7,c=3n=b±b24ac2a=(7)±(7)24(2)(3)2(2)=7±49244=7±254

The radicand in this question (the number under the root) is a square number! That means the answers will be rational numbers, because the square root will tidy up very nicely.

It is a good idea at this point to write the two different answer separately.

n=7±54=7+54and =754=124and =24

The final answers are n=3 and n=12.


Submit your answer as:

Quadratic equations: standard form

Given the quadratic equation:

x2+5x+13=2x23x+3

Which of the following shows the equation in standard form?

Answer:
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Standard form for a quadratic equation looks like this:

0=ax2+bx+c or ax2+bx+c=0

STEP: Rearrange the terms to put the equation in standard form
[−2 points ⇒ 0 / 2 points left]

"Standard form" is a certain arrangement of the equation: there must be a zero on one side and the other terms must be arranged in descending order:

0=ax2+bx+c or ax2+bx+c=0

We can choose if we want to move the terms to the left side of the equation or the right side of the equation. However, it is almost always best to make the quadratic coefficient positive. In this case, that means we should move the terms onto the left side of the equation:

x2+5x+13=2x23x+3x2+8x+10=0

In fact there are many ways to arrange the equation in standard form. For example, another possible answer is:

x28x10=0

One correct answer (and probably the best answer!) is: x2+8x+10=0.


Submit your answer as:

Quadratic equations: standard form

Given the quadratic equation:

5k2+5k+7=1

Which of the following shows the equation in standard form?

Answer:
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Standard form for a quadratic equation looks like this:

0=ax2+bx+c or ax2+bx+c=0

STEP: Rearrange the terms to put the equation in standard form
[−2 points ⇒ 0 / 2 points left]

"Standard form" is a certain arrangement of the equation: there must be a zero on one side and the other terms must be arranged in descending order:

0=ax2+bx+c or ax2+bx+c=0

We can choose if we want to move the terms to the left side of the equation or the right side of the equation. However, it is almost always best to make the quadratic coefficient positive. In this case, that means we should move the terms onto the left side of the equation:

5k2+5k+7=15k2+5k+6=0

In fact there are many ways to arrange the equation in standard form. For example, another possible answer is:

5k25k6=0

One correct answer (and probably the best answer!) is: 5k2+5k+6=0.


Submit your answer as:

Quadratic equations: standard form

Given the quadratic equation:

1=3x2+8x

Which of the following shows the equation in standard form?

Answer:
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Standard form for a quadratic equation looks like this:

0=ax2+bx+c or ax2+bx+c=0

STEP: Rearrange the terms to put the equation in standard form
[−2 points ⇒ 0 / 2 points left]

"Standard form" is a certain arrangement of the equation: there must be a zero on one side and the other terms must be arranged in descending order:

0=ax2+bx+c or ax2+bx+c=0

We can choose if we want to move the terms to the left side of the equation or the right side of the equation. However, it is almost always best to make the quadratic coefficient positive. In this case, that means we should move the terms onto the right side of the equation:

1=3x2+8x0=3x2+8x1

In fact there are many ways to arrange the equation in standard form. For example, another possible answer is:

3x28x+1=0

One correct answer (and probably the best answer!) is: 3x2+8x1=0.


Submit your answer as:

7. Practical applications

Exercises